Non-Perturbative Superpotentials in Landau-Ginzburg Compactification

Abstract:
  We study the Landau-Ginzburg models which correspond to Calabi-Yau
four-folds. We construct the index of the typical states which correspond to
toric divisors. This index shows that whether a corresponding divisor can
generate a non-perturbative superpotential. For an application, we consider the
phase transition in terms of the orbifold construction. We obtain the simple
method by which the divisor, which can not generate a superpotential in the
original theory, can generate a superpotential after orbifoldization.

\section{Introduction}
Recently the Calabi-Yau\ four-folds
have been paied attention to
in the context of
M-theory \cite{witten1,schwarz}
and F-theory
\cite{mv} compactification.
Witten found \cite{witten}
the interesting property of
the Calabi-Yau\ four-fold compactification,
i.e. the special class of the divisors with wrapped 5-branes
can generate the $N = 1$ superpotential.
The condition whether or not a divisor ${\cal D}$
can genarete a superpotential
is purely topological,
i.e. the Euler character
$\chi ({\cal D}) = 1$ is necessary
(not sufficient) and
the holomorphic Hodge numbers
$h^{1,0}({\cal D}) =
h^{2,0}({\cal D})=
h^{3,0}({\cal D}) = 0$
are sufficient
(not necessary) conditions.

In this note,
we study  the such topological properties of divisors
in terms of Landau-Ginzburg\ models.
It is well known that
the Landau-Ginzburg\ models as well as their orbifolds
with central charge $c = 3d$
can describe the Calabi-Yau\ manifolds of dimension $d$.
For example, the Hodge numbers $h^{i,j}$
can be identified with
the numbers of the $(c,c)$ or $(a,c)$ states
with $U(1)$ charge
$(d-i,j)$ or $(-i,j)$ respectively,
where $c$ $(a)$ denotes the (anti-)chiral ring
of $N = 2$ conformal field theory
\cite{lvw}.
These numbers can be calculated
without any geometrical information
and the simple formulas were obtained \cite{sa1}
for three-fold case.
Since four-folds are of our interest,
we will consider the Landau-Ginzburg\ models (and their orbifolds)
with $c = 12$.

For geometrical description of the Calabi-Yau manifold s,
toric geometry is most useful.
In the work of
Klemm, Lian, Roan and Yau \cite{klry},
the topological properties for the divisors
is described in terms of toric geometry.
It is of our interest
to find the corresponding result
in terms of Landau-Ginzburg\ models.
To do this,
we will use the method obtained in \cite{sa3,sa4}
where it was shown how these two description,
the toric geometry and the Landau-Ginzburg\ models,
can be identified.
Once we obtain the Landau-Ginzburg\ description,
we can find the simple conditions for the states
which correspond to the divisors.
These conditions, as expected,
can be applied straightforwardly to the calculation
without any geometrical information.
Moreover, by the orbifold construction of \cite{sa4},
we can easliy find
the new model
in which
the divisor, which can not generate the superpotential
in the original theory,
{\it can} generate the superpotential
in the orbifoldized theory.
This implies that the phase transition
in terms of our orbifoldization
can be characterized by the property of the divisors
which lead to
the non-perturbatively generated superpotentials.
Similar transitions are discussed in \cite{bls}.

This papre is organized as follows.
In section 2, we briefly review the results of \cite{klry}
and their Landau-Ginzburg\ interpretaion is obtained in section 3.
Moreover in section 3, we develop the simple method
to dicide whether or not
a divisor with wrapped 5-brane
can generate a superpotential.
In the last section
we consider the Landau-Ginzburg\ orbifolds,
which correspond to the Calabi-Yau\ four-fold orbifolds,
and their phase transitions.
We can easily find the appropriate orbifoldization
by the method obtained in \cite{sa4},
such that
the divisor which can not generate the superpotential
in the original theory
can generate the superpotential
in the orbifoldized theory.

\section{Toric description of Calabi-Yau\ four-folds
and their divisors}
The topological numbers for Calabi-Yau\ four-folds
are studied in \cite{SethiVafaWitten,klry,Mohri}.
At first sight, the non-trivial Hodge numbers are
$h^{1,1}$, $h^{3,1}$, $h^{2,1}$ and $h^{2,2}$
(up tp dualities).
However by the index theorem \cite{SethiVafaWitten}
it is shown that there is one relation
\begin{equation}
h^{2,2} = 2(22 + 2 h^{1,1} + 2 h^{3,1}  -  h^{2,1}),
\end{equation}
so that only three Hodge numbers are independent,
namely $h^{1,1}$, $h^{3,1}$ and $h^{2,1}$.
It will be clear that
this fact is important
when one considers the relation
between the Landau-Ginzburg\ descripton
and the toric geometry.
The Euler number can be simply written as
\begin{equation}
\chi = 6 (8 +  h^{1,1}  +  h^{3,1}  -  h^{2,1} ).
\end{equation}

There are formulas for these Hodge numbers
$h^{1,1}$, $h^{3,1}$ and $h^{2,1}$
in terms of toric geometry.
The formulas of the first two of these
are obtained
due to Batylev's original work \cite{vb}:
\begin{equation}
\label{vb11}
h^{1,1} =
l(\Delta^{*}) - 6
- \sum_{\bf{codimension} \; {\Theta^{*}} = 1}
{l^{'}({\Theta^{*}})}
+\sum_{\bf{codimension} \; \Theta^{*} = 2,
\Theta^{*} \in \Delta^{*}}
{l^{'}(\Theta^{*})
l^{'}(\Theta)}
,
\end{equation}
\begin{equation}
\label{vb31}
h^{3,1} =
l(\Delta) - 6
- \sum_{\bf{codimension} \; {\Theta} = 1}
{l^{'}({\Theta})}
+ \sum_{\bf{codimension} \; \Theta = 2,
\Theta \in \Delta}
{l^{'}(\Theta)
l^{'}(\Theta^{*})}
,
\end{equation}
where
${l({\Delta})}$ ($l(\Delta^{*})$)
denotes the number of integral points
in the Newton polyhedron ${\Delta}$
(the dual polyhedron ${\Delta}^{*}$)
and
${l^{'}({\Theta})}$ ($l^{'}(\Theta^{*})$)
denotes the number of integral points
interior of the face ${\Theta}$
(the dual face ${\Theta}^{*}$).
The formula for $h^{2,1}$ is obtained
in \cite{klry,Mohri} to be
\begin{equation}
\label{vb21}
h^{2,1} =
\sum_{\bf{codimension} \; \Theta = 3,
\Theta \in \Delta}
{l^{'}(\Theta)
l^{'}(\Theta^{*})}.
\end{equation}

The authors of ref. \cite{klry} studied
the topological numbers of the divisors in Calabi-Yau\ four-folds
using toric geometry.
They analyze the local structure of the divisors
coming from the blowing up of the singularities
on the hypersurface of the Calabi-Yau\ embedded in
the weighted complex projective space.
They have classified the divisors
by thier topological numbers.
We will briefly review their results
which we need.
\begin{description}
\item[Case A] ${d_{{\Theta_{k}^{*}}} = 3}$, \ \
 $h^{0,0}({\cal D}_{k}) = l^{\prime}(\Theta_{k}) + 1$. \ \
 This case explains the additional fourth therm in (\ref{vb11}).
\item[Case B] ${d_{{\Theta_{k}^{*}}} = 2}$, \ \
 $h^{0,0}({\cal D}_{k}) = 1$, \
 $h^{1,0}({\cal D}_{k}) = l^{\prime}(\Theta_{k})$. \ \
 In this case we get
${l^{\prime}(\Theta_{k}^{*})} \cdot {l^{\prime}(\Theta_{k}^{*})}$
$(3,2)$ forms, their dual $(1,2)$, $(2,3)$ and $(2,1)$ forms
on a four-fold,
where we have used the Poincar\'{e}\ and complex cojugation dualities.
\item[Case C] ${d_{{\Theta_{k}^{*}}} = 1}$, \ \
 $h^{0,0}({\cal D}_{k}) = 1$, \
 $h^{2,0}({\cal D}_{k}) = l^{\prime}(\Theta_{k})$. \ \
 On a four-fold, we get additional $(3,1)$ forms
of the fourth term in (\ref{vb31}).
\end{description}
and other $h^{i,j}({\cal D}_{k}) = 0$
in all the cases.
As a result, they find the simple formula for the Euler number
of the divisor ${\cal D}_{k}$, i.e.
\begin{equation}
\label{chi}
\chi ({\cal D}_{k}) =
1 +  (-1)^{{\rm dim}{{\Theta}_{k}} + 1}
\ l^{\prime} ({{\Theta}_{k}}),
\end{equation}
where $l^{\prime} ({{\Theta}_{k}})$ is the number of integral points

in the face ${{\Theta}_{k}}$
which is dual to the point in the dual polyhedron
$\Delta^{*}$ corresponding to
the divisor ${\cal D}_{k}$  of our interest.

\section{Landau-Ginzburg\ analysis}
A Landau-Ginzburg\ model is characterized by a superpotential
$W(X_{i})$ where $X_{i}$ are $N = 2$ chiral superfields.
The Landau-Ginzburg\ orbifolds
\cite{iv} are obtained by
quotienting with an Abelian symmetry group $G$ of
$W{(X_{i})}$,
whose element $g$ acts as an $N \times N$ diagonal matrix,
$g: X_{i} \rightarrow e^{2 \pi i {\tilde{\theta_{i}}^{g}}}X_{i}$,
where $0 \leq \tilde{\theta_{i}}^{g} < 1$.
Of course the $U(1)$ twist \
$j: X_{i} \rightarrow e^{2 \pi i {q_{i}}}X_{i}$ \
generates the symmetry group of
$W{(X_{i})}$,
where $q_{i} = {w_{i} \over d}$,\ \
 $W(\lambda^{w_{i}} X_{i}) = \lambda^{d}W(X_{i})$ \ and
$\lambda \in \hbox{\,$\inbar\kern-.3em{\rm C}$}^{\ast}$.
Using the results of Intriligator and Vafa \cite{iv},
we can construct the $(c,c)$ and $(a,c)$ rings,
where $c$ ($a$) denotes chiral (anti-ciral).
Also we could have the left and right $U(1)$ charges of
the ground state \
$\keta h$ \
in the $h$-twisted sector
of the $(a,c)$ ring.
In terms of spectral flow, \ $\keta h$ is mapped to the (c,c)
state $\ketc {h^{\prime}}$ with $h^{\prime} = hj^{-1}$.
Then the charges of the (a,c) ground state of $h$-twisted sector
$ \keta {h} $
are obtained to be
\begin{equation}
\label{uac}
\begin{array}{cc}
\left(\begin{array}{c}
J_{0} \\
\bar{J_{0}}
\end{array} \right) &
\end{array}
\keta {h}
=
\begin{array}{cc}
\left(\begin{array}{c}
{ - \sum_{\tilde{\theta_{i}}^{h^{\prime}}>0}{(1-q_{i}-\tilde{\theta_{i}}^{h^{\prime}})}}
+ \sum_{\tilde{\theta_{i}}^{h^{\prime}}=0} {(2q_{i}-1)}
 \\
{ \sum_{\tilde{\theta_{i}}^{h^{\prime}}>0}{(1-q_{i}-\tilde{\theta_{i}}^{h^{\prime}})}}
\end{array} \right) &
\keta {h}.
\end{array}
\end{equation}

Our purpose of this section is
to consider the topological properties
of the divisors in terms of the Landau-Ginzburg\ model.
As we reviewed in the previous section,
there are three classes of divisors
of our interest
which are called Case A, B and C.
What is the corresponding classification
in the Landau-Ginzburg\ description?
To answer this question,
we should recall the observation
in ref.\cite{sa4}
where the Calabi-Yau\ three-folds are considered
in terms of the Landau-Ginzburg\ model.

The observation of \cite{sa4} is
that if the $(-1,1)$ state
$\keta {h^{\prime}}$ exists
in the ${h^{\prime}}$-twisted sector
then it is possible to exist
the states written in the form
$\prod_{\tilde{\theta_{i}}^{h^{\prime}} = 0}{X_{i}^{l_{i}}} \keta h$
in the $h$-twisted sector,
where $h^{\prime} \equiv hj^{-1}$.
The $U(1)$ charge of the state
$\prod_{\tilde{\theta_{i}}^{h^{\prime}} = 0}{X_{i}^{l_{i}}} \keta h$
depends on
the number of invariant fields
under the $h^{\prime}$ action.
We denote by $I_{h^{\prime}}$
that number of invariant fields.
For three-folds,if $I_{h^{\prime}} = 2, \  3$,
then the possible $U(1)$ charges are
$(-1,1)$, $(-2,1)$, respectively.

Applying this obsevation to our four-fold case,
we obtain the following results
which correspond to
the previous classification of divisors.
\begin{description}
\item[$I_{h^{\prime}} = 2$] It is possible to exist
$(-1,1)$ states $X_{i}^{l_{i}}X_{j}^{l_{j}} \keta h$.
This case corresponds to the Case A.
\item[$I_{h^{\prime}} = 3$] It is possible to exist
$(-2,1)$ states $X_{i}^{l_{i}}X_{j}^{l_{j}}X_{k}^{l_{k}} \keta h$.
This case corresponds to the Case B.
\item[$I_{h^{\prime}} = 4$] It is possible to exist
$(-3,1)$ states
$X_{i}^{l_{i}}X_{j}^{l_{j}}X_{k}^{l_{k}}X_{m}^{l_{m}} \keta h$.
This case corresponds to the Case C.
\end{description}
where at least one $l_{i} > 0$.
Off course, we have assumed
the existence of the $(-1,1)$ state
$\keta {h^{\prime}}$.
The well-known identification
between $(i,j)$ forms and $(-i,j)$ states
implies
that we can classify
by the number $I_{h^{\prime}}$
the $(-1,1)$ states
which correspond to the divisors.
We will study each case more in depth
and find the condition of the $(-1,1)$ states
whether the corresponding divisors
can generate the superpotntial.

First, consider the case of $I_{h^{\prime}} = 2$.
In this case, all the divisors can generate
the superpotentials.
Since it is shown \cite{sa3,sa4} that
a $(-1,1)$ state $\keta {h^{\prime}}$
corresponds to a integral point
in the dual polyhedron $\Delta^{*}$,
the number of $(-1,1)$ states
$X_{i}^{l_{i}}X_{j}^{l_{j}} \keta h$
in the $h$-twisted sector is equal to
$l^{\prime}(\Theta)$.

Let us turn to the $(c,c)$ ring by spectral flow.
For the four-fold case,
the states which correspond to the $(1,1)$ forms
have thier $U(1)$ charge being $(3,1)$.
Using the technique developed in \cite{sa1},
we obtain the index
$\beta (h^{\prime})$
which counts the number of $(3,1)$ states
written in the form
$X_{i}^{l_{i}}X_{j}^{l_{j}} {\ketc {h^{\prime}}}$,
i.e.
\begin{equation}
\label{beta}
\beta (h^{\prime}) =
\frac{1}{| \rm G |}
\sum_{ {\rm all \ g \ \in \ G}}{
\prod_{ \tilde{\theta_{i}}^{g} = \tilde{\theta_{i}}^{h^{\prime}} = 0}{
(1 - \frac{1}{q_{i}})}
}.
\end{equation}
(If the set of $i$'s satsfying
$\tilde{\theta_{i}}^{g} = \tilde{\theta_{i}}^{h^{\prime}} = 0$ is empty,
then we define
$\prod_{ \tilde{\theta_{i}}^{g} = \tilde{\theta_{i}}^{h^{\prime}} = 0}{
(1 - \frac{1}{q_{i}})}
 = 1$
).
Clearly, in this case we have
$\beta (h^{\prime}) =
l^{\prime}({{\Theta}_{k}})$.

The above discussions can be extended to
other $(-1,1)$ states $\keta {h^{\prime}}$
with 3 or 4 invariant fields under
$h^{\prime}$ action.
Remember that if there are 3 (4) invariant fields
under $h^{\prime}$ action,
it is possible to exist $(-2,1)$ ($ \ (-3,1) \ $) states
$\prod_{\tilde{\theta_{i}}^{h^{\prime}} = 0}{X_{i}^{l_{i}}}{\keta h}$.
For the case with
$I_{h^{\prime}} = 4$,
the index $\beta (h^{\prime})$ is defined
as in (\ref{beta})
over the $(c,c)$ ring
in the $h^{\prime}$-twisted sector.
However in the case with
$I_{h^{\prime}} = 3$,
the index $\beta(h^{\prime})$
should be defined to be
\begin{equation}
\label{beta2}
\beta (h^{\prime}) =
\frac{-1}{2}
\frac{1}{| \rm G |}
\sum_{ {\rm all \ g \ \in \ G}}{
\prod_{ \tilde{\theta_{i}}^{g} = \tilde{\theta_{i}}^{h^{\prime}} = 0}{
(1 - \frac{1}{q_{i}})}
},
\end{equation}
since in the same twisted sector
there are the $(-2,1)$ and $(-1,2)$ states
which corresponds to
the $(2,1)$ and $(1,2)$ forms
obeying the complex conjugation
duality.

Finally we obtain
\begin{equation}
\label{betal}
\beta (h^{\prime}) =
l^{\prime} ({{\Theta}_{k}}),
\end{equation}
in all the cases with
$I_{h^{\prime}} = 2, \ 3, \ 4$.
Note that
${{\rm dim}{{\Theta}_{k}}+ 1} =
I_{h^{\prime}}$.
Comparison of (\ref{chi}) with (\ref{betal})
leads us to conclude that
a non-perturbative superpotential can be generated if
\begin{equation}
\beta (h^{\prime}) = 0.
\end{equation}
This condition is very useful
because the calculation
can be done straightforwardly without
any knowledge of geometry.
Moreover, our condition can be available
for the Landau-Ginzburg\ orbifolds which correspond to
the Calabi-Yau\ four-fold orbifolds.
If one needs to know the toric data
for the divisor
which can genarate the superpotential,
one has only to use the techniques
obtained in \cite{sa3,sa4}.
We will consider the illustrative examples
in the next section.

\section{Applications}
As a first example, we consider the Landau-Ginzburg\ model
which corresponds to
the hypersurface embedded in $WCP_{(1,1,1,1,4,4)}[12]$
\begin{equation}
W_{1} =
X_{1}^{12} +
X_{2}^{12} +
X_{3}^{12} +
X_{4}^{12} +
X_{5}^{3} +
X_{6}^{3} .
\end{equation}
This model is already studied in \cite{witten,mayr}.
By our Landau-Ginzburg\ analysis, we first find two $(-1,1)$ states
$\keta {j^{-1}}$
and
$\keta {j^{-3}}$.
The state $\keta {j^{-1}}$
corresponds to the canonical divisor.
We should pay attention to the fact
that only two fields $X_{5}$ and $X_{6}$
are invariant under $j^{-3}$ action.
So the state $\keta {j^{-3}}$ correponds to
the Case A in the previous section
and the index of (\ref{beta}) is obtained to be
$\beta(j^{-3}) = 2$
(note that the index $\beta$ is calculated
over the $(c,c)$ ring).
This implies there are more two $(-1,1)$ states,
namely
${X_{5}}{\keta {j^{-2}}}$ and
${X_{6}}{\keta {j^{-2}}}$.
For divisors this implies
$h^{0,0}({\cal D}) = 3$,
i.e. three independent divisors intersect
the Calabi-Yau\ hypersurface in the same way \cite{mayr}.

To resolve this unsatisfactory situation,
we apply the orbifold construction developped in \cite{sa4}.
We orbifoldize this model by the $\ZZ_{3}$ twist
$g = {\rho_{5}^{2}}{\rho_{6}}$,
where ${\rho_{i}}{X_{j}} =
{e^{2 \pi i {q_{i}} {\delta_{i,j}} }}{X_{j}}$.
The orbifoldized potential is obtained to be
\begin{equation}
W_{1}^{\prime} =
{X^{\prime}}_{1}^{12} +{X^{\prime}}_{2}^{12}+
{X^{\prime}}_{3}^{12} +{X^{\prime}}_{4}^{12} +
{{X^{\prime}}_{5}^{2}} {{X^{\prime}}_{6}} +
{{X^{\prime}}_{5}} {{X^{\prime}}_{6}^{2}}.
\end{equation}
The two $(-1,1)$ states
${X_{5}}{\keta {j^{-2}}}$ and
${X_{6}}{\keta {j^{-2}}}$
in the original theory
are mapped to in the orbifoldized theory
$\keta {{g^{-1}}{j^{-2}}}$ and
$\keta {{g^{-2}}{j^{-2}}}$,
respectively.
Applying the method in \cite{sa4},
the toric data for the $(-1,1)$ state
${\keta {j^{-3}}}$,
$\keta {{g^{-1}}{j^{-2}}}$ and
$\keta {{g^{-2}}{j^{-2}}}$
are obtained to be
$(0,0,0,-1,-1)$,
$(0,0,0,1,0)$ and
$(0,0,0,0,1)$,
respectively.
Thus the three divisors,
which can not have the toric data independently
in the original theory,
are represented independently
in the orbifoldized theory.
All of these divisors with wrapped 5-branes
can generate superpotentials.
These two theories of $W_{1}$ and ${W^{\prime}}_{1}$
are conected by this orbifold transition
as was shown in \cite{sa4}.

Our next example is the Landau-Ginzburg\ model whose potential is
\begin{equation}
W_{2} =
X_{1}^{24} +
X_{2}^{24} +
X_{3}^{12} +
X_{4}^{6} +
X_{5}^{3} +
X_{6}^{3},
\end{equation}
which corresponds to the hypersurface
embedded in $WCP_{(1,1,2,4,8,8)}[24]$.
This is the three-fold fibered Calabi-Yau\ four-fold
\cite{bs}
and its Hodge numbers are
$h^{1,1} = 6$,
$h^{3,1} = 803$,
$h^{2,1} = 1$ and
$h^{2,2} = 3,278$.

We will concentrate on the pair of states
$\keta {j^{-6}}$ of $U(1)$ charge $(-1,1)$ and
${X_{4}}{\keta {j^{-5}}}$ of $(-2,1)$,
since this pair corresponds to the Case B
in the previous section.
Thus the divisor corresponding to
the $(-1,1)$ state $\keta {j^{-6}}$
cannot generate a superpotential.
The index of (\ref{beta2}) is obtained to be
$\beta(j^{-6}) = 1$,
as expected.
The toric data of
the $(-1,1)$ state
$\keta {j^{-6}}$ is obtained to be
$(0,0,-1,-2,-2)$.

We will show that
after taking an appropriate orbifoldization
this divisor can generate a superpotential.
If we apply the appropriate orbifoldization,
then the $(-2,1)$ state
${X_{4}}{\keta {j^{-5}}}$
is projected out,
so that the divisor corresponds to
the $(-1,1)$ state
$\keta {j^{-6}}$
{\it can} generate the superpotential.
We can easily find such an appropriate orbifoldization
by the method obtained in \cite{sa4}.

In this case,
we should consider the $\ZZ_{3}$ orbifoldization
by the twist
$g = {{\rho_{4}^{2}}{\rho_{5}}{\rho_{6}}}$.
In the orbifoldized theory,
the $(-2,1)$ state
${X_{4}}{\keta {j^{-5}}}$
is projected out
and
the index $\beta(j^{-6})$ is calculated to be
$\beta(j^{-6}) = 0$
as expected.
Thus the divisor corresponds to
the $(-1,1)$ state
$\keta {j^{-6}}$
{\it can} generate a superpotential.
As was shown in \cite{sa4},
the original and the orbifoldized theories
are connected.
This phase transition can be characterized
by the property of the divisors
which lead to
the non-perturbatively generated superpotentials.

As usual, new twisted states will appear
in the orbifoldized theory.
For our model,
two new $(-2,1)$ states
$\keta {{g^{-1}}{j^{-23}}}$ and
$\keta {{g^{-2}}{j^{-23}}}$
arise.
So the new divisors which correspond to
the new $(-1,1)$ states
$\keta {{g^{-1}}}$ and
$\keta {{g^{-2}}}$
cannot generate the superpotentials.

It is interesting to note that
the $(-2,1)$ state
${X_{4}}{\keta {j^{-5}}}$
in the original theory
is mapped to
the $(-1,1)$ state
${\keta {{g^{-1}}{j^{-5}}}}$
which can generate the superpotential.
In general,
by the $U(1)$ charge analysis of \cite{sa4},
we see that the $(-p,1)$ states
in the original theory
written in the form
$\prod_{\tilde{\theta_{i}}^{h^{\prime}} = 0}{X_{i}^{l_{i}}} \keta h$
for $p = 1, \ 2, \ 3$
can be mapped to the $(-1,1)$ states
$\keta {{g^{-l}}{h}}$
in our orbifoldized theory.
In other words,
if there exists
one divisor in the original four-fold
which cannot generate the superpotential,
we obtain several divisors
in the orbifoldized four-fold,
at least one of which
can generate a superpotential.

\vspace{1cm}

{\it Acknowledgements} :
The author would like to thank H. Itoyama
for valuable discussions
and his continuous encouragement.

\newpage


On the Scalar Curvature of Einstein Manifolds

Abstract:
  We show that there are high-dimensional smooth compact manifolds which admit
pairs of Einstein metrics for which the scalar curvatures have opposite signs.
These are counter-examples to a conjecture considered by Besse. The proof
hinges on showing that the Barlow surface has small deformations with ample
canonical line bundle.

\section{Introduction}

Let $M$ be a smooth compact $n$-manifold. An {\em Einstein
metric} on $M$ is a smooth Riemannian metric $g$ on
$M$ such that
$$r =\lambda g,$$
where $r $ is the Ricci curvature of $g$ and
$\lambda$ is any real  constant. Such a
metric  has scalar curvature $s =n\lambda$, so
the   so-called Einstein constant $\lambda$
 and the scalar curvature
of $g$ have the same sign.

When $n < 4$, any Einstein metric has constant
sectional curvature $\lambda /(n-1)$. In dimensions
$2$ and $3$, the sign of $\lambda$ is therefore
determined by the size of the fundamental group,
and so is really a topological invariant of $M$.
This motivated Besse \cite[p.19]{bes} to consider
the conjecture that  no smooth compact $n$-manifold can ever admit
Einstein metrics with different signs of $\lambda$.

The K\"ahler-Einstein case superficially
seems to support such a conjecture.
 Indeed, if $J_1$ and $J_2$ are
deformation-equivalent complex structures
on an even dimension smooth manifold $M$,
then
there cannot be a K\"ahler-Einstein metric with $\lambda > 0$
compatible with $J_1$ and a K\"ahler-Einstein metric with
$\lambda < 0$ compatible with $J_2$.
When $n=4$, Seiberg-Witten theory \cite{witten,leb}
and the Hitchin-Thorpe inequality \cite{hit}
allow one to make an even more encouraging
statement: if a smooth compact 4-manifold   admits a
 K\"ahler-Einstein metric
with $\lambda \leq 0$, then it cannot admit  {\em any}
Einstein metric
for which  $\lambda$ has a different sign.

 Unfortunately, these pieces of
 evidence are just red herrings.
Not only is the conjecture false, but
counter-examples can be constructed as K\"ahler-Einstein
metrics on products of complex surfaces!

\begin{thm} For
any $k \geq 2$, there is a  smooth compact $4k$-manifolds $M$
which  admits both an Einstein metric $g_1$ with $\lambda > 0$
and an Einstein metric $g_2$ with $\lambda < 0$.
Moreover, $M$ may be taken to be the $k$-fold product
of ${\bf CP}_2\#8\overline{\bf CP}_2$ with itself,
and $g_1$ and $g_2$ may both be taken to be K\"ahler-Einstein
metrics.
\end{thm}

It follows, of course, that the relevant complex structures
    $J_1$ and $J_2$ on $M$ are
deformation-inequivalent.

 The    proof begins with the observation that
${\bf CP}_2\# 8 \overline{\bf CP}_2$
is $h$-cobordant to the   complex surface  $S$ of general type
discovered by Barlow \cite{bar}.
While
Barlow's explicit complex structures on $S$ do not have
ample canonical bundle,  we show in \S\S \ref{first}--\ref{sec}
that suitable small deformations of these  complex structures
do indeed have
$c_1 < 0$. The theory the complex Monge-Amp\`ere equation
\cite{aubin,tian,ty,yau}
and standard facts about $h$-cobordisms \cite{smale,wall}
  then imply that  Einstein's equation can be solved  on
$S\times S$   both when  $\lambda >0$
and when $\lambda < 0$.

Incidentally, the low-dimensional failure of the $h$-cobordism
theorem is illustrated  by the
fact that  $S$ and ${\bf CP}_2\# 8 \overline{\bf CP}_2$ are themselves
non-diffeomorphic \cite{kot,ov}; cf. \cite{leb2}.
On the other hand,
Freedman's
work \cite{f} tells us that
just enough of the $h$-cobordism theorem survives
to conclude that the Einstein manifolds
 $S$ and ${\bf CP}_2\# 8 \overline{\bf CP}_2$ are {\em homeomorphic}.
Thus, while the Besse conjecture still stands in dimension 4,
it survives only by virtue of the existence of
exotic differentiable structures on 4-manifolds.

 So far as we know, these are also the first examples
of smooth compact manifolds which admit
both K\"ahler metrics of
positive Ricci curvature and
K\"ahler metrics of  negative Ricci curvature.
However,   Lohkamp \cite{loh}, generalizing an earlier result of
Gao and Yau \cite{gao},
 has shown that absolutely every
smooth  manifold of dimension $n > 2$ admits
Riemannian metrics of negative
Ricci curvature, so it is only the
K\"ahler condition that gives this last  observation
 any new interest.  Rethinking the 3-dimensional
case in light of these last results should
serve to remind the reader  that the
 Einstein condition  is immeasurably
 stronger than the mere requirement
  that the
Ricci curvature have a fixed sign.

\section{Deformations of the Barlow Surface}
\label{first}

Let $F$ be a quintic surface in ${\bf CP}_3$ with
exactly 20 nodes and no other singularities.
Each of the  nodes is locally
modeled by ${\bf C}^2/{\bf Z}_2$,
since the quadratic
map ${\bf C}^2\to \odot^2{\bf C}^2: v\mapsto v\otimes v$
identifies ${\bf C}^2/{\bf Z}_2$ with the
nodal surface $\det =0$ in $\odot^2{\bf C}^2\cong {\bf C}^3$.
Thus
$F$  naturally carries the structure of a complex orbifold.
We will be interested in the case in which
$F$ is a global orbifold:
$$F= Y/{\bf Z}_2$$
for some compact complex 2-manifold $Y$
with  a holomorphic involution. Using the
theory of Hilbert modular surfaces, one can \cite{cat}, for example,
 show that this is true of
the specific 20-nodal quintic
surface
\begin{eqnarray} \sum_{j=1}^5 z_j^5&=&\frac{5}{4}
 (\sum_{j=1}^5 z_j^2)( \sum_{j=1}^5 z_j^3) \label{symm}\\
\sum_{j=1}^5 z_j &=& 0 \nonumber ,\end{eqnarray}
 in a hyperplane of ${\bf CP}_4$.
For more details, see \S \ref{sec}.

We will need two basic facts about the surface $Y$.
The first of these is rather obvious:

\begin{lem}
In the above situation, $Y$ has ample canonical line
bundle.
\end{lem}
\begin{proof}
The canonical line bundle
bundle of $Y$ is the pull-back of ${\cal O}(1)$
from ${\bf CP}_3$ via the tautological map
$Y\to F$. Since this map is finite-to-one,
it does not collapse any
curves. The canonical line bundle is therefore
positive by Nakai's criterion.
\end{proof}

The second basic  fact is, by contrast, quite delicate:
\setcounter{bis}{\thethm}

\begin{thm}\label{terza}
In the above  situation, $H^2(Y,TY)=0$.
\end{thm}
\noindent
In the interest of clarity, we defer the rather long proof
to \S \ref{sec}.

\medskip

 Now consider the particular quintic $F$ given by
equation \ref{symm}. The symmetric group ${\cal S}_5$ acts
on $F$, and in particular we have an action of
the dihedral group $D_{10}\subset {\cal S}_5$
 of pentagonal isometries, generated by
$(25)(34)$ and $(12345)$. This action lifts
to $Y$ in such a way that \cite{cat} the cyclic subgroup
${\bf Z}_5\subset D_{10}$ generated by
$(12345)$ acts freely on $Y$ and such that \cite{bar}
the involution
$(25)(34)$ acts with exactly 20 fixed   points.
The so-called Catanese surface $X=Y/{\bf Z}_5$
is therefore
non-singular, and comes equipped with an
involution $\alpha : X\to X$ with exactly
4 fixed points. From the above basic facts
we immediately read off

\begin{lem}
The Catanese surface $X$ has ample canonical
bundle and satisfies $H^2(X,TX)=0$.
\label{cs}
\end{lem}
\begin{proof} These follow from general facts about
unramified covers. For example,
since $-c_1(Y)$ is represented by a positive
2-form, it is also, by averaging, represented by
a positive ${\bf Z}_5$-invariant 2-form;
such a form descends to $X$, and represents $-c_1(X)$.
In the same vein, the pull-back maps such as
$H^2(X,TX)\to H^2(Y,TY)$ are necessarily injective,
since the pull-back of a harmonic representative
is harmonic with respect to an invariant metric.
\end{proof}

Dividing $X$ by the action of the
the involution $\alpha$, we obtain
a surface $X/{\bf Z}_2$ whose
only singularities are four nodes.
The {\em Barlow surface} is by definition
   the minimal resolution  $S$
of  $X/{\bf Z}_2$. One can show \cite{bar}
that  $S$ is a minimal,
simply connected  complex
surface of general type,
with
$c_1^2=1$ and $p_g=0$.

By construction, $S$ contains four
$(-2)$-curves $B_1, \ldots , B_4$,
arising from the nodes of $X/{\bf Z}_2$;
in particular, $S$ does {\em not} have
ample canonical bundle. On the other hand,

\begin{lem} The only  $(-2)$-curves in $S$ are
  $B_1, \ldots , B_4$.
\end{lem}
\begin{proof}
Because  the orbifold $X/{\bf Z}_2$ has $K$ ample,
 the underlying variety has  a projective embedding
in projective space \cite{bail} which lifts to a
pluri-canonical map of $S$ collapsing only the
$B_j$.
\end{proof}

Let us  use $B$ to  denote the union of
the four $(-2)$-curves $B_1, \ldots , B_4$
in $S$, and let $TS (- \log B)$ denote the
sheaf of vector fields on $S$ with trivial
normal component along the curve $B$.

\begin{lem} $H^2(S, TS (- \log B))=0$.
\label{bs}
\end{lem}
\begin{proof}
 The Serre dual of $H^2(S, TS )(- \log B)$
is $H^0(S, \Omega^1( \log B) \otimes \Omega^2)$.
 The latter  is  \cite[Prop. 3.1]{cat2} exactly the
${\bf Z}_2$-invariant subspace of
$H^0(\tilde{X}, \Omega^1 \otimes \Omega^2)=
H^0({X}, \Omega^1 \otimes \Omega^2)$.
However, the latter is Serre dual to
$H^2(X, TX)$, which vanishes by Lemma \ref{cs}.
\end{proof}

\begin{thm} The Barlow surface $S$ has a
  smooth versal deformation space
of complex dimension 8. A general
point in this space corresponds to a surface
with ample canonical bundle.
\label{key}
\end{thm}
\begin{proof}
Let $\nu$ denote the normal bundle of
$B$. We then have an exact sequence
$$0\to  TS(- \log B)\to  TS \to {\cal O}_B(\nu)\to 0$$
of sheaves on $S$. Thus Lemma \ref{bs}
tells us, in particular, that
$H^2(S,TS)=0$.
 Kodaira-Spencer theory thus
gives us a smooth versal deformation of
$S$ with tangent space $H^1(S,TS)$, the dimension of which is
$h^1(S,TS)=-\chi (S, TS) =  (5c_2-7c_1^2)/6= (5\cdot 11 - 7\cdot 1)/6=
8.$ (Note that $H^0(S,TS)=0$ because
$S$ is of general type.)

Now for each $j=1, \ldots , 4$, define $TS (-\log B_j)$ to be the
sheaf of holomorphic vector fields with trivial
normal component along the curve $B_j$. We then
have   exact sequences
$$0\to TS(- \log B)\to TS(-\log B_j)
\to {\cal O}_{
\bar{B}_j}(\nu)\to 0,
$$
where $\bar{B}_j=\cup_{j\neq k}B_k$
is the complement of $B_j\subset B$.
Thus
 the natural maps $H^1(S, TS (-\log B_j))\to
H^1(S,TS)$ have images of codimension 1.
Choose a curve through the base-point of  the Kodaira-Spencer
family, parameterized by an embedding of the unit disk
$\Delta\subset {\bf C}$.
Let ${\cal S}\to \Delta$ be the induced family
with central fiber $S$. By construction, the Kodaira-Spencer
map of this family at zero does not take values in
$H^1(S,TS (-\log B_j))$ for any $j\in \{1, 2,3 , 4\}$.
Now   there are only finitely
many classes   $a\in H^2(S, {\bf Z})$ such that  $c_1\cdot a=0$
and  $a\cdot a=-2$, since $c_1^2> 0$ and
the intersection form is of Lorentz type. Since
$q(S)=0$, each such $a$ corresponds to a unique
holomorphic line bundle $L_{a,t}\to S_t$
for each $t\in \Delta$, where $S_t$ is the fiber over $t$.
We have $h^0(S_t, L_{a,t})\leq 1$ for all $(a,t)$, since
$(-2)$ curves are necessarily isolated, so
 semi-continuity \cite[\S 10.5]{gr} implies that, for each $a$,
 the set of $t\in \Delta$
where $h^0(S_t,L_{a,t}) = 1$ is closed in the
analytic Zariski topology. Now suppose there were such
an $a$
such that $0\in \Delta$ wasn't isolated in
$\{ t\in \Delta ~|~ h^0(S_t,L_{a,t}) = 1\}$.
Then we would have $h^0(S_t,L_{a,t}) = 1$
$\forall t\in \Delta$, and the  direct
image theorem \cite[\S 10.5]{gr}
would assert the existence
of a   section of $L_{t,a}$ which is non-trivial for each $t$ and
depends holomorphically on $t$. Thus there
would be a family of $(-2)$-curves in
the fibers of ${\cal S}\to \Delta$ depending
smoothly on the base, and in particular
 it would follow that $a=[B_j]$ for
some $j$. Moreover, we would have a
family of curves $B_{j,t}\subset S_t$ depending
holomorphically on $t$, so that ${\cal S}$ actually
comes from a deformation of the pair $(S,B_j)$.
But  its Kodaira-Spencer map
at the base-point would therefore take values in
$H^1(S, TS (-\log B_j))$, in contradiction to our
assumptions. This shows that there is a non-empty
subset of points in the Kodaira-Spencer space
corresponding to surfaces without $(-2)$-curves;
and by semi-continuity, this non-empty subset
is the complement of an analytic subset. It follows
\cite{bpv}
that the general point of the Kodaira-Spencer space
corresponds to a surface with ample canonical bundle.
\end{proof}

For the experts, the  above elementary argument may be considerably
shortened  by
invoking the results of  \cite{bw,cat3}. Indeed,
from this point of view, Lemma \ref{bs}
asserts
that there is a smooth submersion
$\mbox{Def }({X/{\bf Z}_2})\stackrel{\psi}{\to} {\cal L}_{X/{\bf Z}_2}$,
where ${\cal L}_{X/{\bf Z}_2}$ is the local
deformation space of the four nodes
of ${X/{\bf Z}_2}$. Thus by \cite[(1.8)]{cat3},
we have a fiber product of smooth manifolds
$$
\begin{array}
{ccc}
\mbox{Def }(S) &\to &{\cal L}_S\\
\downarrow &&
\downarrow_{\beta}\\
\mbox{Def }({X/{\bf Z}_2})&\stackrel{\psi}{\to} &{\cal L}_{X/{\bf Z}_2}
\end{array}
$$
where $\beta : {\bf C}^4\to {\bf C}^4$
is given by $(z_j)\mapsto (z_j^2)$. Thus
a general deformation of $S$ has a smooth canonical
model --- i.e. has ample canonical bundle.

\section{Vanishing Theorems for Nodal Surfaces}
\label{sec}

In this section, we will prove  the key vanishing result
Theorem \ref{terza}. We will do this
by  first proving some
general results concerning the following   class of
nodal surfaces:

\begin{defn}
Let $x\mapsto [a_{ij}(x)]$ be a linear map from
${\bf C}^4$ to the symmetric $n\times n$ matrices,
and let $F\subset {\bf CP}_3$ denote the
 surface of degree $n$ defined by $\det ([a_{ij}(x)])=0$.
Suppose that the only singular points at which
$\mbox{\rm rank} [a_{ij}(x)] \leq n-2$
 are nodes.
Then we shall say that $F$ is a {\em nodal linearly
symmetric surface} of degree $n$.
\end{defn}

It turns out that there are precisely
 ${n+1\choose 3}$ nodes \cite{cat}, which we shall
denote by  $p_j\in F$,
$j=1,\ldots,  {n+1\choose 3}$. Let $\tilde{F}$
be the smooth surface obtained by blowing up these nodes,
let $\pi : \tilde{F}\to F$ be the blowing-down map, and set
$A_j=\pi^{-1}(p_j)$. The $A_j$ are thus disjoint $(-2)$-curves,
and we denote their union by $A=\cup_j A_j$.
Setting ${\cal O}_{\tilde{F}}(H)= \pi^*{\cal O}_F(1)$ and
  $r=[\frac{n}{2}]$, there is then a divisor $L$ on
$\tilde{F}$ such that
$2L=A+ (2r+1-n)H$. It was  shown in
\cite[Theorem 2.19]{cat}
that $h^0(\tilde{F}, {\cal O}(rH-L))=n$, and
the base locus of the system $|rH-L|$ is empty.
We thus have a holomorphic map
$\varepsilon : \tilde{F}\to {\bf CP}_3\times {\bf CP}_{n-1}$
given by $|H|\times |rH-L|$, and $\varepsilon$ is an embedding,
since $|H|$ alone suffices away from the nodes and
$|rH-L|$ is a degree-$1$ embedding on each $A_j$.
 It was also  shown that, relative to a suitable basis
$v_j$, $j=1,\ldots , n$ for  $H^0(\tilde{F}, {\cal O}(rH-L))$
that the image of $\varepsilon$ is exactly given by
the equations
$$\sum_{j=1}^n a_{ij}(x)v_j =0, ~i=1,\ldots , n.$$
As was pointed out by L. Ein, this proves the following
useful fact:

\begin{lem} \label{cap}
The blow-up $\tilde{F}$ of such a surface $F$ of degree $n$
at its nodes
is a
complete intersection of $n$ hypersurfaces in
${\bf CP}_3\times {\bf CP}_{n-1}$ of bidegree
$(1,1)$.
\end{lem}
In particular, if $\tilde{F}$ is smooth,
its   normal bundle in ${\bf CP}_3\times {\bf CP}_{n-1}$
is just the restriction of
$[{\cal O}(1,1)]^{\oplus n}$ to $\tilde{F}$.
We also get
the following useful vanishing result:

\begin{lem}
\label{zip}
If $n > 4$, then
$$ H^1 (\tilde{F}, {\cal O}(a,b)) =0$$
provided
that
\begin{itemize}
\item  $a,b\geq -1$; or
\item $b < 2$, $a < n-2$; or
\item $-1\leq a \leq n-3$; or
\item $-1\leq b\leq 1$.
\end{itemize}
Moreover,
$$H^0({\bf CP}_3\times {\bf CP}_{n-1},
{\cal O}(a,b))\longrightarrow H^0(\tilde{F},
{\cal O}(a,b))$$
is surjective provided
that
\begin{itemize}
\item $a,b\geq 0$; or
\item $b < 3$, $a < n-1$; or
\item $0\leq a \leq n-2$; or
\item $0\leq b\leq 2$.
\end{itemize}
\end{lem}
\begin{proof}
Let $E=[{\cal O}_{{\bf CP}_3\times
{\bf CP}_{n-1}}(-1,-1)]^{\oplus n}$.
Because $\tilde{F}$ is a complete
intersection, we have the
Koszul complex
$$
0\to \Lambda^nE\to \cdots \to \Lambda^2E\to E\to
{\cal O}_{{\bf CP}_3\times
{\bf CP}_{n-1}}\to
{\cal O}_{\tilde{F}}\to 0,$$
and we may tensor this with ${\cal O}(a,b)$ to obtain an
exact complex
$$0\to {\cal E}^{-n}\to \cdots \to {\cal E}^{-2}\to {\cal E}^{-1}
\to {\cal O}(a,b)\to {\cal O}_{\tilde{F}}(a,b)\to 0.$$
Since a sum of line bundles on a projective space can only have
non-trivial cohomology in the top or bottom dimension,
 the K\"unneth formula tells us that this complex always satisfies
 $H^i({\cal E}^{-j})=0$ when $i\not\in\{ 0,3,n-1 , n+2\}$.
By breaking the complex into  short exact sequences, it thus follows
that $H^1(\tilde{F}, {\cal O}(a,b))=0$
whenever
 $H^3({\cal E}^{-2})=H^{n-1}({\cal E}^{-(n-2)})=0$.
 But ${\cal E}^{-2}$ is a direct sum of copies of
${\cal O}(a-2,b-2)$, so that $H^3({\cal E}^{-2}) =0$
provided that $a > -2$ or $b < 2$. Similarly,
since ${\cal E}^{-(n-2)}$ is a direct sum of copies of
${\cal O}(a-(n-2),b-(n-2))$, we have
$H^{n-1}({\cal E}^{-(n-2)})=0$ provided
that $a < n-2$ or $b > -2$.

The same splicing argument
shows that the surjectivity of
$$H^0({\bf CP}_3\times {\bf CP}_{n-1},
{\cal O}(a,b))\longrightarrow H^0(\tilde{F},
{\cal O}(a,b))$$
 follows whenever
$H^3({\cal E}^{-3})=H^{n-1}({\cal E}^{-(n-1)})=0$.
Since ${\cal E}^{-3}$ and ${\cal E}^{-(n-1)}$
are respectively   sums of copies of
${\cal O}(a-3,b-3)$
and ${\cal O}(a-(n-1),b-(n-1))$,
one has $a > -1$ or $b < 3 ~\Rightarrow H^3({\cal E}^{-3})=0$,
and $a < n-2$ or $b > -1  ~\Rightarrow H^{n-1}({\cal E}^{-(n-1)})=0$.
\end{proof}

This, incidentally, yields a sharpening
 of  \cite[Proposition 2.25]{cat}:

\begin{cor}
Let $p_j\in {\bf CP}_3$ be the node points of $F$, as before,
and for each $j=1, \ldots , {n+1\choose 3}$, consider the
condition of passing through $p_j$
as a linear constraint on the linear system
of surfaces of degree $(n-1)$ in ${\bf CP}_3$.
Then these ${n+1\choose 3}$ conditions are linearly independent.
\end{cor}
\begin{proof}
The restriction map
$$H^0({\bf CP}_3\times {\bf CP}_{n-1}, {\cal O}(0,2))\to
H^0(\tilde{F}, {\cal O}(0,2))$$
is surjective by Lemma \ref{zip},
and it is injective by the same reasoning. However,
${\cal O}_{\tilde{F}}((n-1)H-A)={\cal O}_{\tilde{F}}(0,2)$
by construction.
Thus
$$h^0(\tilde{F}, {\cal O}((n-1) H-A))=
h^0({\bf CP}_3\times {\bf CP}_{n-1}, {\cal O}(0,2))=
{n+1\choose 2}.$$
But the restrict-and-pull-back
 map $H^0({\bf CP}_3, {\cal O}(n-1))\to
H^0(\tilde{F}, {\cal O}((n-1) H))$ is also an isomorphism,
so $H^0(\tilde{F}, {\cal O}((n-1) H-A))$ can be
identified with the subspace of
$H^0({\bf CP}_3, {\cal O}(n-1))$ consisting of
sections which vanish at the $p_j$. Since
$${
n+1 \choose 2}=
{
n+2\choose 3}-
{n+1\choose 3}=
h^0({\bf CP}_3, {\cal O}(n-1)) - \# \{ p_j\}, $$
the claim therefore follows.
\end{proof}

We now specialize to the $n=5$ case. Let
$F$ be a linearly symmetric quintic whose only
singularities are its 20 nodes;  an example
is \cite{cat} given by \ref{symm}.  Thus $\tilde{F}$
is smooth. Moreover, $A=2L$, so there is a double
branched cover $\tilde{Y}$ ramified exactly over the
exceptional curves $A_j$. The inverse image of
each $A_j$ is an $(-1)$-curve $E_j$, and we
may blow these 20 disjoint $(-1)$-curves down to
obtain a smooth surface $Y$, which comes equipped with
a natural involution $\Phi: Y\to Y$
corresponding to the
sheet-interchanging map $\tilde{\Phi}: \tilde{Y}\to
\tilde{Y}$  of $\tilde{Y}\to \tilde{F}$.
We thus have $F=Y/{\bf Z}_2$, realizing $F$ as a
global orbifold, exactly as claimed in \S \ref{first}.

We are now prepared to prove the key vanishing result
used in the previous
section:

\begin{bis}
Let $Y$ be as above. Then $H^2(Y,TY)=0$.
\end{bis}
\begin{proof}
The Serre dual of $H^2(Y,TY)$ is $H^0(Y, \Omega^1\otimes \Omega^2)$,
and the latter can be identified with
$H^0(\tilde{Y}, \Omega^1\otimes \Omega^2)$
as a consequence of  Hartogs' theorem. Let us express
the latter in terms of the $(\pm 1)$-eigenspaces
of the action of $\tilde{\Phi}$:
$$H^0(\tilde{Y}, \Omega^1\otimes \Omega^2)
=H^0(\tilde{Y}, \Omega^1\otimes \Omega^2)^+\oplus
H^0(\tilde{Y}, \Omega^1\otimes \Omega^2)^-.$$
By standard formulae \cite[Prop. 3.1]{cat2},
these eigenspaces have direct interpretations on $\tilde{F}$:
\begin{eqnarray*}
 H^0(\tilde{Y}, \Omega^1\otimes \Omega^2)^+&=&
 H^0(\tilde{F}, \Omega^1(\log A)\otimes \Omega^2)
\\
H^0(\tilde{Y}, \Omega^1\otimes \Omega^2)^-&=&
H^0(\tilde{F}, \Omega^1 \otimes \Omega^2\otimes L).
\end{eqnarray*}
In fact, however, by \cite[Proposition 1.6]{cat3},
the first statement has an ostensible
  strengthening:
$$H^0(\tilde{Y}, \Omega^1\otimes \Omega^2)^+=
H^0(\tilde{F}, \Omega^1 \otimes \pi^*\omega_F) =
H^0(\tilde{F}, \Omega^1 \otimes \Omega^2) .$$
This strengthening is Serre dual to
the equality
$$H^2(\tilde{F}, T\tilde{F}(-\log A))=
H^2(\tilde{F}, T\tilde{F}),$$
which follows directly \cite[Corollary 1.3]{bw}
from the observation that ${F}$ has deformations which
smooth all its nodes.

Let us now compute $H^0(\tilde{Y}, \Omega^1\otimes \Omega^2)^+=
H^0(\tilde{F}, \Omega^1 \otimes \Omega^2\otimes L)$.
By Lemma \ref{cap}, $\tilde{F}$ is the complete
intersection of five hypersurfaces in ${\bf CP}_3\times
{\bf CP}_4$ of bidegree $(1,1)$. The canonical line
bundle of $\tilde{F}$ is therefore
$$\Omega^2_{\tilde{F}}= {\cal O}_{\tilde{F}}(-4,-5)\otimes
{\cal O}_{\tilde{F}}(5,5)= {\cal O}_{\tilde{F}}(1,0),$$
whereas
$${\cal O}(L)= {\cal O}(2H) \otimes
[{\cal O}(2H-L)]^*  = {\cal O}_{\tilde{F}}(2,-1).$$
Thus our objective is to show that
$H^0( \Omega^1_{\tilde{F}} (3,-1))$ vanishes.
Because $H^1(\tilde{F}, {\cal O}(2,-2))=0$
by  Lemma \ref{zip}, tensoring the conormal
bundle sequence
$$0\to [{\cal O}_{\tilde{F}} (-1,-1)]^{\oplus 5} \to
 \hat{\Omega}^1
\to \Omega^1_{\tilde{F}}\to 0$$
with ${\cal O}(3,-1)$
shows that it is enough to ascertain the
vanishing of
$H^0(\tilde{F},  \hat{\Omega}^1 (3,-1) )$,
where   $\hat{\Omega}^1=\Omega^1_{{\bf P} \times{\bf P}}
\otimes {\cal O}_{\tilde{F}}$
is the restriction of the cotangent bundle of
${\bf CP}_3\times{\bf CP}_4$ to $\tilde{F}$.
However,   tensoring the
Euler exact sequence
$$0\to \hat{\Omega}^1 \to
[{\cal O}(-1,0)]^{\oplus 4}\oplus [{\cal O}(0,-1)]^{\oplus 5}\to
{\cal O}\oplus {\cal O}\to 0$$
with ${\cal O}_{\tilde{F}}(3,-1)$ tells us that
$$  H^0(\tilde{F}, \hat{\Omega}^1 (3,-1)
 )\subset
[H^0(\tilde{F}, {\cal O}(3,-2))]^{\oplus 4}\oplus
[H^0(\tilde{F}, {\cal O}(2,-1))]^{\oplus 5} .$$
On the other hand,  $H^0({\bf CP}_3\times {\bf CP}_4,
{\cal O}(3,-2))\to H^0(\tilde{F},
{\cal O}(3,-2))$ and $H^0({\bf CP}_3\times {\bf CP}_4,
{\cal O}(2,-1))\to H^0(\tilde{F},
{\cal O}(2,-1))$ are both surjective by Lemma
\ref{zip}; and since the relevant line bundles on
${\bf CP}_3\times {\bf CP}_4$ obviously do not
admit non-trivial sections, we conclude that
$H^0(\tilde{Y}, \Omega^1\otimes \Omega^2)^-=0$.

To finish the proof, we now compute
$H^0(\tilde{Y}, \Omega^1\otimes \Omega^2)^+=
H^0(\tilde{F}, \Omega^1 \otimes \Omega^2)$
by the same method. Indeed,
$\Omega^1_{\tilde{F}}\otimes \Omega^2_{\tilde{F}}=
 \Omega^1_{\tilde{F}}(1,0)$, and the conormal sequence
tells us that
$$H^0(\tilde{F},  \hat{\Omega}^1(1,0)) \to
H^0(\tilde{F},  \Omega^1(1,0))\to
[H^1 (\tilde{F}, {\cal O}(0,-1))]^{\oplus 5}  $$
is exact. Since  $H^1 (\tilde{F}, {\cal O}(0,-1))=0$ by
Lemma \ref{zip}, we need only verify that
$H^0(\tilde{F},  \hat{\Omega}^1(1,0))$ vanishes, too.
Now the Euler sequence tells us that
$$
H^0(\tilde{F},  \hat{\Omega}^1(1,0))= \ker
\left(
\begin{array}{c}
 [ H^0(\tilde{F},  {\cal O})]^{\oplus 4} \cr
 \oplus  \cr
 [ H^0
( \tilde{F},  {\cal O}(1,-1)) ]^{\oplus 5}
\end{array}
\right) \to
\left( \begin{array}{c}
H^0 (\tilde{F},
{\cal O}(1,0)) \\
\oplus \\
 H^0 (\tilde{F},  {\cal O}(1,0))
\end{array}
\right)   .
$$
However, $H^0(\tilde{F},  {\cal O}(1,-1))=0$
by Lemma \ref{zip}, and $[H^0(\tilde{F},  {\cal O})]^{\oplus 4}\to
H^0(\tilde{F},  {\cal O}(1,0))$ is manifestly injective.
Thus
$H^0(\tilde{Y}, \Omega^1\otimes \Omega^2)^+=0$,
and the result follows.
\end{proof}

It is worth noting that
the above result depends heavily on the assumption that
$F$ is quintic. Indeed, one can use  much the same argument
to show that the
the  vanishing assertion definitely  fails iff $n \gg 5$.

\section{Einstein Manifolds and Smooth Topology}

We begin by recalling that two smooth compact
oriented $n$-manifolds are said to be
$h$-cobordant \cite{smale} if there is a compact oriented
$(n+1)$-manifold $W$ whose boundary is $M\cup \bar{N}$,
and such that the inclusion of either boundary component
is a homotopy equivalence; here $\bar{N}$ denotes the
orientation-reversed version of $N$. One then says
that $W$ is an $h$-cobordism from $M$ to $N$.
This defines an equivalence relation on the
set of all $n$-manifolds, since gluing two $h$-cobordisms
end-to-end yields a new $h$-cobordism.
This equivalence relation is also compatible with
Cartesian products:

\begin{lem}
Let $M$ and $N$ be $h$-cobordant   manifolds.
 Then their Cartesian self-products
$M\times M$ and $N\times N$ are also
$h$-cobordant. Moreover, the
 iterated self-products
$M^{(k)}=M\times \cdots \times M$ and
$N^{(k)}=N\times \cdots \times N$ are
$h$-cobordant for any $k\geq 1$.
\end{lem}
\begin{proof}
If $W$ is a an $h$-cobordism between $M$ and $N$,
then $M\times W$ is an $h$-cobordism between $M\times M$ and
$M\times N$, whereas $W\times N$
is an $h$-cobordism between $M\times N$ and
$N\times N$; gluing these end-to-end then
gives the desired $h$-cobordism.

The iterated case follows   similarly, using
induction. Namely, if $W_{k-1}$
is an $h$-cobordism from $M^{(k-1)}$ to
$N^{(k-1)}$, then $M\times W_{k-1}$
is an $h$-cobordism between $M^{(k-1)}$  and
 $M\times N^{(k-1)}$,
whereas $W\times N^{(k-1)}$ is an
$h$-cobordism between $M\times N^{(k-1)}$
and $N^{(k)}$.
\end{proof}

This observation allows one to prove the following useful fact:

\begin{prop}
Let $M$ and $N$ be two  smooth simply connected
compact 4-manifolds
with equal Euler characteristics
$\chi (M)=\chi (N)$ and   signatures
$\tau (M)=\tau (N)\not\equiv 0 \bmod 16$.
Then $M\times M$ is diffeomorphic to $N\times N$. Moreover,
the   iterated Cartesian self-products $M^{(k)}$
and $N^{(k)}$ are diffeomorphic
for any $k > 1$.
\end{prop}
 \begin{proof}
By Rochlin's theorem \cite{ah}, $M$ and $N$ are not spin, and so have
odd intersection form. Since $b_2(M)=b_2(N)$
and $\tau (M)=\tau (N)$, the intersection forms of
$M$ and $N$ are therefore isomorphic by the
Minkowksi-Hasse classification \cite{hm} and Donaldson's
thesis \cite{don}.
A  theorem of Wall \cite{wall}
therefore tells us that $M$ and $N$ are
$h$-cobordant. By the previous lemma,
it follows that
of $M^{(k)}$ and $N^{(k)}$ are $h$-cobordant for any
$k$. But for any $k > 1$, these
are simply connected compact smooth manifolds
of dimension $> 5$, so Smale's $h$-cobordism theorem
  asserts that they are diffeomorphic, as claimed.
\end{proof}

\begin{cor}
Let $S$ denote the Barlow surface, and let $R$
denote the rational complex surface ${\bf CP}_2\#
8 \overline{\bf CP}_2$.
Then $S^{(k)}$ is diffeomorphic to $R^{(k)}$
for all $k > 1$. In particular, $S\times S$ is diffeomorphic
to $R\times R$.
\end{cor}
\begin{proof}
Both $R$ are $S$ are simply connected complex surfaces with
$c_1^2=1$ and $p_g=0$. Thus both have $\chi =11$ and
$\tau = -7 \not\equiv 0 \bmod 16$. The claim thus follows
from the preceding result.
\end{proof}

\begin{thm}
Let $R={\bf CP}_2\#
8 \overline{\bf CP}_2$. Then the
8-manifold $R\times R$ admits Einstein
metrics of both positive and negative scalar
curvatures. Similarly, the $4k$-manifold
$R^{(k)}$ admits Einstein
metrics of both positive and negative scalar
curvatures for all $k > 1$. Moreover, the
relevant
Einstein metrics can be constructed so
as to be K\"ahler with respect to suitable
complex structures.
\end{thm}
\begin{proof}
By Theorem \ref{key}, the Barlow surface $(S,J_0)$ has
deformations with ample canonical line
bundle.
If $J_t$ is any complex structure on $S$ with this
property,   the work of Aubin and Yau \cite{aubin,yau}
tells us that there is a unique Einstein metric $g_S$
on $S$ which is K\"ahler with respect to $J_t$ and
has scalar curvature $-1$. On the other hand,
if $\tilde{J}$ is a complex  structure
on $R$ corresponding to a  blow-up of
${\bf CP}_2$ at 8 points in general position,
then Tian \cite{tian}, building on his joint
work with Yau \cite{ty}, has shown that
$R$ admits an Einstein metric $g_R$ of scalar
curvature $+1$ which is K\"ahler with respect
to $\tilde{J}$.

Now because the Ricci tensor of any Riemannian
product is the direct  sum of the Ricci tensors
of the factors, any Riemannian product of two
Einstein manifolds with the same value of $\lambda$
is again Einstein. In particular, Cartesian self-products
of Einstein manifolds are always Einstein, and, moreover,
Cartesian self-products of K\"ahler-Einstein manifolds
are always K\"ahler-Einstein. Thus $R^{(k)}$
admits K\"ahler-Einstein metrics with
positive scalar curvature, and $S^{(k)}$
admits K\"ahler-Einstein metrics with
negative scalar curvature. But by the above Corollary,
these manifolds are diffeomorphic, and the relevant
Einstein metrics may therefore both be
considered as living on $R^{(k)}$.
\end{proof}

\bigskip

\noindent
{\bf Acknowledgements.}
The first author would like to
thank L. Ein for pointing out that $\tilde{F}$
is a complete intersection. He would also like to
to thank P. Supino for his queries regarding Theorem
\ref{terza}.


JM Figueroa-O'Farrill, C Koehl and B Spence

Supersymmetry and the cohomology of (hyper)Kaehler manifolds

15 pages (LaTeX2e) (A factor has been corrected in one equation.)

Abstract:
  The cohomology of a compact Kaehler (resp. hyperKaehler) manifold admits the
action of the Lie algebra so(2,1) (resp. so(4,1)). In this paper we show,
following an idea of Witten, how this action follows from supersymmetry, in
particular from the symmetries of certain supersymmetric sigma models. In
addition, many of the fundamental identities in Hodge-Lefschetz theory are also
naturally derived from supersymmetry.

\section{Introduction}

It is a classical result going back to Hodge and Lefschetz
\cite{GriffithsHarris} that the de~Rham cohomology of a compact
K\"ahler manifold admits an action of the Lie algebra $\mathrm{s}\ell(2) \cong
\mathrm{so}(2,1)$.  This action is generated by the operations of exterior
product with the K\"ahler form and its adjoint operation with respect
to the Hodge inner product.  A more recent result of Verbitsky
\cite{Verbitsky-so(5),Verbitsky} states that if the manifold is
hyperK\"ahler, then the $\mathrm{so}(2,1)$ action is part of a larger $\mathrm{so}(4,1)$
action, which is now generated by exterior products with each of the
three K\"ahler forms and their adjoints.  Recently Witten
\cite{WittenTalk} has suggested that this can be understood from
supersymmetry.

It is well-known that requiring supersymmetry in a sigma model imposes
strong restrictions on the geometry of the target manifold $X$.  The
earliest result in this direction was the observation \cite{Zumino}
that the four-dimensional sigma model can be made supersymmetric if
and only if $X$ admits a K\"ahler metric.  Similarly
\cite{AlvarezGaumeFreedman} supersymmetry of a six-dimensional sigma
model with target space $X$ demands that $X$ be hyperK\"ahler.  Upon
dimensional reduction to one dimension these supersymmetric sigma
models become quantum mechanical systems, introduced to great effect
in \cite{WittenIndex} and much studied since.  These quantum
mechanical systems have as Hilbert space the square-integrable
differential forms on $X$, and as hamiltonian the Hodge laplacian.  By
reason of supersymmetry, the energy is non-negative.  Therefore the
ground states are in one-to-one correspondence with the
square-integrable harmonic forms and thus, when $X$ is compact, with
the cohomology.  Therefore any symmetry of the hamiltonian will
preserve the ground states and hence act on the cohomology.  On the
other hand, a $(d{+}1)$-dimensional supersymmetric sigma model with a
Lorentzian metric has an action which is invariant under the Lorentz
group $\mathrm{SO}(d,1)$.  Upon dimensional reduction to one (spatial)
dimension, this reduces to an $\mathrm{SO}(d{-}1,1)$ `internal' symmetry.
Applying this to supersymmetric sigma models in $3{+}1$
(resp. $5{+}1$) dimensions with compact target space $X$ yields an
$\mathrm{SO}(2,1)$ (resp. $\mathrm{SO}(4,1)$) action on the cohomology of $X$.  This
is Witten's argument.

Whereas this beautiful argument explains the existence of these group
actions, it does not tell us that they agree with the ones known from
geometry.  It is the purpose of this paper to prove that they do.  In
fact, supersymmetry lies at the heart of much of the Hodge--Lefschetz
theory and in this paper we will also briefly mention these results.
For example, the Hodge identities are what remains (after dimensional
reduction) of the fact that the supersymmetry generators are spinors of
the relevant (i.e., four- or six-dimensional) spacetime.

This paper is organised as follows.  In the next section we will
consider the K\"ahler case, reviewing the Hodge--Lefschetz
construction and then checking explicitly that it agrees with the one
coming from supersymmetry.  Then in section 3 we will tackle the
hyperK\"ahler case, reviewing Verbitsky's construction and checking
that it too agrees with that coming from supersymmetry.  Finally in
Section 4 we offer some closing comments.

\section{The K\"ahler case}

In this section we discuss the K\"ahler case.  We first briefly review
some facts from the Hodge--Lefschetz theory of a K\"ahler manifold which
we will need. We will then show how these may be understood
by considering the dimensional reduction of
the four-dimensional supersymmetric sigma model.
We follow the notational conventions of
\cite{WessBagger}.

\subsection{Some harmonic theory on K\"ahler manifolds}

Let $X$ be a compact K\"ahler manifold of complex dimension $n$.
The complex differential forms on $X$ decompose as
\begin{equation*}
A_c(X) = \bigoplus_{p,q=1,\ldots,n} A_c^{p,q}~,
\end{equation*}
where $A_c^{p,q}$ is the subspace of $(p,q)$-forms.
The Hodge $\star$-operator maps
\begin{equation*}
\star: A_c^{p,q} \to A_c^{n-q,n-p}~,
\end{equation*}
and obeys $\star^2 = (-)^{p+q}$ acting on $A_c^{p,q}$.  Using this
operator we may define an hermitian inner product on the space of
forms, the Hodge inner product:
\begin{equation}\label{eq:hodgeip}
(\alpha,\beta) = \int_X \alpha\wedge\star\Bar\beta~.
\end{equation}
Notice that because complex conjugation maps $A_c^{p,q} \to
A_c^{q,p}$, the Hodge inner product pairs forms of the same bidegree.

The de~Rham operator on $X$ decomposes into two operators of different
bidegree: $d = \d + \Bar\d$, where $\d : A_c^{p,q} \to A_c^{p+1,q}$
and $\Bar\d : A_c^{p,q} \to A_c^{p,q+1}$.  $\Bar\d$ is called the
Dolbeault operator.  From $d^2=0$, one reads that $\d^2 = \Bar\d^2 =
0$ and $\d\Bar\d = - \Bar\d\d$.  Let $\d^*$ and $\Bar\d^*$ denote
their adjoint operators relative to the Hodge inner product.  They
also satisfy similar identities.  Using these operators we can define
two laplacians $\square$ and $\Bar\square$ as follows:
\begin{equation*}
\square = \d\d^* + \d^* \d \qquad \text{and}\qquad
\Bar\square = \Bar\d\Bar\d^* + \Bar\d^* \Bar\d~.
\end{equation*}
Part of the magic of K\"ahler manifolds is that these two laplacians
agree.  In fact, we have
\begin{equation}\label{eq:laplacians}
\square = \Bar\square = \tfrac{1}{2} \bigtriangleup~,
\end{equation}
where $\bigtriangleup$ is the Hodge laplacian.  Other identities
obeyed by the differential operators $\d$, $\Bar\d$, $\d^*$, and
$\Bar\d^*$ are:
\begin{equation}\label{eq:more-D-ids}
\d\Bar\d^* = - \Bar\d^*\d \qquad\text{and}\qquad
\d^*\Bar\d = - \Bar\d\d^*~.
\end{equation}

Now let $\omega \in A_c^{1,1}$ denote the K\"ahler form.  It is real:
$\Bar\omega = \omega$.  Let $L: A_c^{p,q} \to A_c^{p+1,q+1}$ be
defined by $L(\alpha) = \alpha \wedge\omega$.  Its adjoint relative to
the Hodge inner product is denoted $\Lambda: A_c^{p,q} \to
A_c^{p-1,q-1}$, and obeys the relations
\begin{equation*}
(\Lambda(\alpha),\beta) = (\alpha, L(\beta)) = (\alpha, \beta \wedge
\omega)~.
\end{equation*}
It can be understood as contraction with the K\"ahler form.  Let us
now define a third operator $H \equiv [L,\Lambda] : A_c^{p,q} \to
A_c^{p,q}$.  Actually $H$ acts diagonally with real eigenvalue $p+q-n$
on $A_c^{p,q}$.  Furthermore, the following relations hold:
\begin{equation*}
{}[H,L] = 2L \qquad\text{and}\qquad [H,\Lambda] = -2\Lambda~.
\end{equation*}
In other words, $L$, $\Lambda$ and $H$ define a representation of the
Lie algebra $sl(2)$.

These operators obey a series of identities, known as the Hodge
identities, which relate the Dolbeault operator and its cousins.
These identities read:
\begin{alignat}{4}\label{eq:Hodgeids}
[L,\d] &= 0 &\quad [L,\Bar\d] &= 0 &\quad [\Lambda,\d^*] &= 0 &\quad
[\Lambda,\Bar\d^*] &= 0\notag\\
[L,\d^*] &= i\Bar\d &\quad [L,\Bar\d^*] &= -i\d &\quad [\Lambda,\d] &=
i\Bar\d^* &\quad [\Lambda,\Bar\d] &= -i\d^*~.
\end{alignat}
It follows from these identities, that $L$, $\Lambda$ and hence $H$
commute with the Hodge laplacian.  This means that they act on the
space of harmonic forms and, by the Hodge decomposition theorem, on
the cohomology.  As we will see below, all the identities in this
section will follow naturally from supersymmetry.

\subsection{The supersymmetric sigma model in four dimensions}

We will now recover these results using supersymmetry.  Let $X$ be a
K\"ahler manifold of complex dimension $n$.  Let $z^a$ and $\Bar
z^{\Bar a}$ denote local complex coordinates, relative to which the
metric has nonzero components $g_{a\Bar b}$.  The nonzero components
of the Christoffel symbols are denoted $\Gamma^a_{bc}$ and
$\Gamma^{\Bar a}_{\Bar b\Bar c}$, and those of the fully covariant
Riemann curvature tensor by $R_{a\Bar b c \Bar d}$.

The $N{=}1$ supersymmetric sigma model in four dimensions with target
space $X$ is described by the following lagrangian density:
\begin{equation}\label{eq:lagr}
\EuScript{L} = -g_{a \Bar{b}} \d_{\mu} \phi^{a} \d^{\mu}
\Bar{\phi}^{\Bar{b}} - \tfrac{i}{2} g_{a \Bar{b}} \Bar{\chi}^{\Bar{b}}
\Bar{\sigma}^{\mu} \overleftrightarrow{D}_{\mu} \chi^{a} +
\tfrac{1}{4} R_{a \Bar{b} c \Bar{d}} (\chi^a\chi^c)
(\Bar\chi^{\Bar b}\Bar\chi^{\Bar d})~,
\end{equation}
where
\begin{itemize}
\item $\phi^a$ is a complex scalar field with $\Bar\phi^{\Bar a} =
(\phi^a)^*$;
\item $\chi^a_\alpha$ is a complex Weyl spinor and $\Bar\chi^{\Bar
a}_{\Dot\alpha} = (\chi^a_\alpha)^*$;
\item $\Bar\sigma^\mu = (-\mathbb{1},-\sigma^i)$ and $\sigma^\mu =
(-\mathbb{1},\sigma^i)$, where $\sigma^i$ are the Pauli matrices; and
\item the covariant derivative is defined by
\begin{equation*}
D_{\mu} \chi^{a} = \d_{\mu} \chi^{a} + \Gamma^{a}_{bc}
\d_{\mu} \phi^{b} \chi^{c}~.
\end{equation*}
\end{itemize}

The lagrangian \eqref{eq:lagr} is invariant under the superpoincar\'e
group, which contains an $\mathrm{SO}(3,1)$ subgroup generated infinitesimally
by:
\begin{align*}
\delta_{\Lambda} \phi^a &= - \Lambda_{\mu\nu} x^\mu
\d^{\nu}\phi^{a}\\
\delta_{\Lambda} \chi^a &= -\Lambda_{\mu\nu} x^\mu \d^\nu
\chi^a + \tfrac{1}{2} \Lambda_{\mu\nu} \sigma^{\mu\nu}\chi^{a}\\
\delta_{\Lambda} \Bar\chi^{\Bar a} &= -\Lambda_{\mu\nu} x^\mu
\d^\nu \Bar\chi^{\Bar a} + \tfrac{1}{2} \Lambda_{\mu\nu}
\Bar\chi^{\Bar a} \Bar\sigma^{\mu\nu}~,
\end{align*}
where $\Lambda_{\mu\nu}$ is constant and antisymmetric.  The Noether
current associated with $\mathrm{so}(3,1)$ transformations can be worked out
by letting the parameter $\Lambda_{\mu\nu}$ depend on position and
varying the lagrangian.
Then up to a total derivative, we find after some $\sigma$-algebra:
\begin{equation}\label{eq:current}
\delta_{\Lambda} \EuScript{L} = \tfrac{i}{2} g_{a\Bar b} \d^\mu
\Lambda_{\mu\nu} \Bar\chi^{\Bar b} \Bar\sigma^\nu \chi^a~,
\end{equation}
where we have omitted terms which will not survive the dimensional
reduction.

The lagrangian \eqref{eq:lagr} is also invariant under the
supersymmetry transformations:
\begin{align*}
\delta_\varepsilon \phi^a &= \varepsilon\chi^a\\
\delta_\varepsilon \chi^a &= i\sigma^\mu \Bar\varepsilon \d_\mu\phi^a
- \Gamma^a_{bc}\delta_\varepsilon\phi^b \chi^c~,
\end{align*}
where $\varepsilon$ is a complex two-component spinor.  The
supersymmetry currents $S^\mu_\alpha$ and $\Bar S^{\mu\,\Dot\alpha}$
can be obtained just as was done for the Lorentz currents, and one
finds:
\begin{equation}\label{eq:susycurrent}
S^\mu = g_{a\Bar b} \d_\nu\Bar\phi^{\Bar b}
\sigma^\nu\Bar\sigma^\mu\chi^a~,
\end{equation}
with
$\Bar S^{\mu\,\Dot\alpha} = \epsilon^{\Dot\alpha\Dot\beta}
\left(S^\mu_\beta\right)^*$.

We now perform a trivial dimensional reduction to one dimension by
simply dropping the dependence on all coordinates but $x^0$.  The
dimensionally reduced lagrangian then becomes:
\begin{equation*}
\EuScript{L} = g_{a\bar b} \Dot\phi^a \Dot{\Bar\phi^{\Bar b}} - \tfrac{i}{2}
g_{a \bar b} \Bar\chi^{\Bar b} \bar\sigma^0
\overleftrightarrow{\frac{D}{dt}} \chi^a +
\tfrac{1}{4} R_{a \Bar{b} c \Bar{d}} (\chi^a\chi^c)
(\Bar\chi^{\Bar b}\Bar\chi^{\Bar d})~,
\end{equation*}
with $\frac{D}{dt} = D_0$.  The conserved charges of the reduced
lagrangian are now
\begin{equation*}
J^{ij} = - {i\over2}g_{a\bar b}\bar\chi^{\bar
b}\bar\sigma^0\sigma^{ij}\chi^a,\qquad\text{with $i,j=1,2,3$,}
\end{equation*}
for the `internal' Lorentz generators, and
\begin{equation*}
S = g_{a\Bar b} \chi^a \Dot{\Bar\phi}^{\Bar b},
\end{equation*}
for the supercharges.

Defining $J_i = \tfrac{1}{2}\epsilon_{ijk}J^{jk}$ and after some
algebra, we find that
\begin{equation}\label{eq:noether}
J^i = \tfrac{i}{2} g_{a \Bar b} \Bar\chi^{\Bar b} \Bar\sigma^i
\chi^a.
\end{equation}

The quantisation of this system is well-known.  The quantisation of
the bosons $\phi^a$ and $\Bar\phi^{\Bar a}$ is straightforward: the
bosonic Hilbert space $\EuScript{B}$ is the space of square integrable
complex-valued functions $f(\phi,\Bar\phi)$.  The canonical
anticommutation relations of the fermions are given by the following
Clifford algebra:
\begin{equation*}
\{\chi^a_\alpha, \Bar\chi^{\Bar b}_{\Dot\alpha}\} = -g^{a\Bar
b}(\phi,\Bar\phi)\sigma^0_{\alpha\Dot\alpha}~.
\end{equation*}
(Note that in our conventions $-\sigma^0_{\alpha\Dot\alpha}$ coincides
with the $2\times2$ identity matrix.)
We can choose a Clifford vacuum $|0\rangle$ by the condition that
$\Bar\chi^{\Bar b}_{\Dot2} |0\rangle = \chi^a_1 |0\rangle = 0$.  Then
a typical state in the fermionic Hilbert space $\EuScript{F}$ is a linear
combination of monomials of the form:
\begin{equation*}
|\psi\rangle = \chi^{a_1}_2\chi^{a_2}_2\cdots\chi^{a_p}_2
\Bar\chi^{\Bar b_1}_{\Dot 1}\Bar\chi^{\Bar b_2}_{\Dot
1}\cdots\Bar\chi^{\Bar b_q}_{\Dot 1} |0\rangle~.
\end{equation*}
More precisely, states like these for fixed $p$ and $q$ generate a
subspace $\EuScript{F}^{p,q}$ of the total fermionic Hilbert space.  The inner
product in $\EuScript{F}$ is defined as follows: if $|\psi\rangle$ is as above,
its norm is given by $\langle\psi|\psi\rangle$ where
\begin{equation*}
\langle\psi| = \langle0| \chi_1^{b_q}\chi_1^{b_{q-1}}\cdots
\chi_1^{b_1} \Bar\chi_{\Dot2}^{\Bar a_p}\Bar\chi_{\Dot2}^{\Bar
a_{p-1}}\cdots\Bar\chi_{\Dot2}^{\Bar a_1}~.
\end{equation*}
Tensoring bosons and fermions together we see that the total Hilbert
space $\EuScript{H} = \EuScript{B} \otimes \EuScript{F}$ is isomorphic to the space of
square-integrable complex differential forms on $X$ relative to the
Hodge metric.  Since $X$ is compact, $\EuScript{H}$ has a dense subspace
isomorphic to the smooth complex forms $A_c(X)$.  Under the
isomorphism, $\EuScript{H}$ inherits a bigrading $\EuScript{H} = \bigoplus \EuScript{H}^{p,q}$
which agrees with the one coming from the fermionic Hilbert space.

Under this isomorphism, operators in the quantum theory can be
interpreted geometrically as operators acting on $A_c(X)$.  It is not
hard to show that the following dictionary holds for the supersymmetry
generators:
\begin{equation}\label{eq:dictionary}
S_1 \mapsto \Bar\d^* \qquad S_2 \mapsto \d \qquad \Bar S_{\Dot1}
\mapsto \Bar\d \qquad \Bar S_{\Dot2} \mapsto \d^*~.
\end{equation}
The supersymmetry algebra obeyed by the supercharges, which can easily
be worked out by iterating the supersymmetry transformations, then
implies the identities \eqref{eq:laplacians} and
\eqref{eq:more-D-ids}.  In particular, this shows that up to an
inconsequential factor, the hamiltonian can be interpreted as the
Hodge laplacian.

Now consider the following linear combination of the Noether charges
\eqref{eq:noether}:
\begin{equation*}
L = J^{1} + i J^{2} = -i g_{a \Bar b}\Bar\chi^{\Bar b}_{\Dot1}\chi^a_2
= \omega_{a\Bar b}\Bar\chi^{\Bar b}_{\Dot1}\chi^a_2~.
\end{equation*}
Under the isomorphism $\EuScript{H} \cong A_c(X)$, we see that $L$ agrees with
the operator $L$ defined in the previous subsection.  Its hermitian
adjoint is clearly given by
\begin{equation*}
\Lambda \equiv L^\dagger = - \omega_{b\Bar a}\Bar\chi^{\Bar a}_{\Dot2}\chi^b_1
= -(J^1 - i J^2)~.
\end{equation*}
Their commutator is given by
\begin{equation*}
H \equiv [L, L^\dagger] = 2 i J^3 = g_{a\Bar b} (\Bar\chi^{\Bar
b}_{\Dot 1} \chi_1^a - \Bar\chi^{\Bar b}_{\Dot 2} \chi_2^a)~,
\end{equation*}
which when written in terms of normal-ordered quantities becomes:
\begin{equation*}
H = g_{a\Bar b} (\Bar\chi^{\Bar b}_{\Dot 1} \chi_1^a +
\chi_2^a \Bar\chi^{\Bar b}_{\Dot 2}) - n\mathbb{1}~,
\end{equation*}
which acts as $(p+q-n)\mathbb{1}$ on $\EuScript{H}^{p,q}$, as expected.  Thus under the
isomorphism $\EuScript{H} \cong A_c(X)$, the operators $(L, \Lambda, H)$ go
over to their namesakes.  Notice that although the $J^i$ satisfy the
Lie algebra of $\mathrm{so}(3)$, the expressions for $L$, $\Lambda$ and $H$
involve complex linear combinations of the $J^i$.  The hermiticity
conditions are such that we are in effect choosing a different real
section of $\mathrm{so}(3,\mathbb{C})$, one isomorphic to $\mathrm{so}(2,1) \cong \mathrm{s}\ell(2)$.
Had we kept $x^1$, say, in the dimensional reduction, we would have
obtained $\mathrm{so}(2,1)$ directly without having to take complex linear
combinations; but such a choice would have made the notation a lot
more complicated.

\begin{table}[h!]
\begin{tabular}{|>{$}l<{$}|>{$}l<{$}|}
\hline
\multicolumn{1}{|c|}{Sigma model} & \multicolumn{1}{c|}{K\"ahler
geometry}\\
\hline\hline
\text{Hilbert space $\EuScript{H} = \bigoplus_{p,q} \EuScript{H}^{p,q}$} & A_c(X) =
\bigoplus_{p,q} A_c^{p,q} \\
\text{inner product} & \text{Hodge inner product \eqref{eq:hodgeip}}\\
\text{supercharge $\Bar S_{\Dot\alpha}$} & (\Bar\d,\d^*) \\
\text{supercharge $S_{\alpha}$} & (\Bar\d^*,\d) \\
\text{hamiltonian} & \bigtriangleup \\
\text{ground states} & H^*(X)\\
\text{`internal' symmetry $J^i$} & L,\,\Lambda,\,H \\
\text{supersymmetry algebra} &
\begin{cases}
\square = \Bar\square = \tfrac{1}{2} \bigtriangleup & \\
\text{$\d$-identities \eqref{eq:more-D-ids}} &
\end{cases}\\
\text{$(S_\alpha, \Bar S^{\Dot\alpha})$ is a spinor in $3{+}1$:
\eqref{eq:susyHodgeids}} & \text{Hodge identities
\eqref{eq:Hodgeids}}\\
\hline
\end{tabular}
\vspace{8pt}
\caption{Sigma models and K\"ahler geometry.\label{tab:susyklr}}
\end{table}

Finally we remark that the Hodge identities \eqref{eq:Hodgeids} are a
consequence of the spinorial nature of the four-dimensional
supercharge.  If we let $\mathbb{S} = (S_\alpha~\Bar S^{\Dot\alpha})^t$
denote the four-dimensional supercharge, then $[J^{\mu\nu},\mathbb{S}] =
\Sigma^{\mu\nu} \mathbb{S}$.  Upon dimensional reduction, we find that
$[J^{ij}, S] = \sigma^{ij}\,S$, or simply that
\begin{equation}\label{eq:susyHodgeids}
[J^i,S] = \tfrac{i}{2} \sigma^i\Bar\sigma^0 S~.
\end{equation}
Using the expressions of $L$ and $\Lambda$ in terms of the $J^i$ and
the dictionary \eqref{eq:dictionary} it is easy to show that
\eqref{eq:susyHodgeids} goes over to the Hodge identities
\eqref{eq:Hodgeids}.  More invariantly, notice that $\mathbb{S}$ is a
Majorana spinor in $3{+}1$ dimensions, hence under $\mathrm{Spin}^0(3,1) \cong
\mathrm{SL}(2,\mathbb{C})$, it transforms according to the real representation
$(0,\half) \oplus (\half,0)$.  Under the `internal' $\mathrm{Spin}^0(2,1)
\cong \mathrm{SL}(2,\mathbb{R})$ subgroup of $\mathrm{Spin}^0(3,1)$ this representation
breaks up into two doublets.  The Hodge identities simply reiterate
this fact. Table~\ref{tab:susyklr} above summarises the correspondence
described in this section.

\section{The hyperK\"ahler case}

Now comes the turn of the hyperK\"ahler case.  As before we first
briefly review Verbitsky's construction of the action of $\mathrm{so}(4,1)$ on
the cohomology of a compact hyperK\"ahler manifold and then show how
this can be reproduced via dimensional reduction of the
six-dimensional supersymmetric sigma model.

\subsection{Verbitsky's construction}

Let $X$ be a compact hyperK\"ahler manifold, and let $I$, $J$ and $K$
be the three complex structures and $\omega_i$, $i=1,2,3$ be the
corresponding K\"ahler forms.  We fix a choice of one of the
complex structures, $I$, say.  Ignoring for the moment the other two
complex structures, $X$ is a compact K\"ahler manifold and all of the
Hodge--Lefschetz theory goes through.  In particular we have operators
$L_1$, $\Lambda_1$ and $H$ defined as before but with $\omega_1$
playing the role of $\omega$.

We now bring to play the other two complex structures.  We define
operators $L_2$ and $L_3$ in the obvious way.  Their adjoints relative
to the Hodge inner product are $\Lambda_2$ and $\Lambda_3$.
Introducing operators $K_i = \tfrac{1}{2} \epsilon_{ijk}
[L_j,\Lambda_k]$, Verbitsky \cite{Verbitsky} showed that the
following algebra is satisfied:
\begin{alignat}{2}
[L_i, \Lambda_j] &= \epsilon_{ijk} K_k + \delta_{ij} H & \qquad
[K_i, K_j] &= \epsilon_{ijk} K_k \notag\\
[H, L_i] & = 2 L_i & \qquad [H, \Lambda_i] & = -2 \Lambda_i
\label{eq:so(4,1)}\\
[K_i,L_j] &= \epsilon_{ijk} L_k & \qquad
[K_i,\Lambda_j] &= \epsilon_{ijk} \Lambda_k\notag
\end{alignat}
with all other brackets zero.  Furthermore, Verbitsky also showed that
as in the K\"ahler case, these operators commute with the Hodge
laplacian, thus inducing an action on the cohomology.

The Lie algebra above is that of $\mathrm{so}(4,1)$. For the purposes of the
calculations in the next section, we will utilise the following
description of these generators.  Let $J_{mn}$, for $m,n =
1,\ldots,5$, denote the generators of $\mathrm{so}(5)$, satisfying the algebra
\begin{equation}\label{eq:so(5)}
[J_{mn} , J_{pq}] =  \delta_{mq} J_{np} - \delta_{mp} J_{nq} +
\delta_{np} J_{mq} - \delta_{nq} J_{mp}~,
\end{equation}
Now, defining (for $i,j = 1,2,3$)
\begin{alignat*}{2}
L_i &= -\left(J_{i5} + iJ_{i4}\right) &\qquad  \Lambda_i &= J_{i5} -
iJ_{i4},\\
H &= 2iJ_{45} &\qquad K_{ij} &= 2J_{ij},
\end{alignat*}
one finds that the algebra of these generators reproduces
\eqref{eq:so(4,1)}.  Again, taking complex linear combinations we have
moved to a different real section of $\mathrm{so}(5,\mathbb{C})$, this time
$\mathrm{so}(4,1)$ as evinced by the presence of factors of $i$ in connection
with the fourth coordinate.

\subsection{The six-dimensional supersymmetric sigma model}

We will now recover the results above using the supersymmetric sigma
model in $5{+}1$ dimensions.  The target space of this sigma model is
a hyperK\"ahler manifold $X$ of (real) dimension $4n$.  Normally the
fermions in the sigma model would be a section of the positive spinor
bundle $S_+$ over spacetime twisted by the pull-back of the tangent
bundle $T$ of the target manifold.  However in this case, this
prescription does not give rise to a match between the bosonic and
fermionic degrees of freedom: we must impose a restriction on the
fermions which we now detail.  The complexified tangent bundle $T_\mathbb{C}$
of a hyperK\"ahler manifold decomposes under the maximal subgroup
$\Sp(1)\cdot \Sp(n) \subset \mathrm{SO}(4n)$ as $T_\mathbb{C} \cong \Sigma \otimes
V$, where $\Sigma$ is a complex two-dimensional $\Sp(1)$ bundle and
$V$ is a complex $2n$-dimensional $\Sp(n)$ bundle.  The holonomy being
$\Sp(n)$ means that the above decomposition is preserved under
parallel transport and that in addition $\Sigma$ is a trivial bundle.
The canonical real structure of $T_\mathbb{C}$ is the product of the natural
quaternionic structures in $\Sigma$ and $V$.  Because $S_+$ also
possesses a quaternionic structure, the tensor products $S_+\otimes V$
and $S_+\otimes\Sigma$ possess real structures.  Therefore we will be
able to impose reality conditions on the fermions and on the
supersymmetry parameters respectively.\footnote{Such spinors are known
as symplectic Majorana--Weyl spinors, and they exist in spacetimes of
signature $(s,t)$ with $s-t=4\mod 8$.}  The bundle $S_+\otimes V$ is
complex $8n$-dimensional.  The reality condition leaves $8n$ real
components which gives $4n$ physical degrees of freedom, matching the
number of bosonic physical degrees of freedom.  Similarly
$S_+\otimes\Sigma$ is complex 8-dimensional and the reality condition
leaves the expected 8 real components of the supercharge.

We now introduce some notation to describe the fields in the sigma
model.  First we have $4n$ bosons $\phi^i$ which are coordinates of the
target manifold.  The isomorphism $T_\mathbb{C} \cong \Sigma\otimes V$ is
given explicitly by objects $\gamma^i_{A a}$.  Here
$A,B,\ldots$ are $\Sp(1)$ indices associated with $\Sigma$
and running from $1$ to $2$, and $a,b,\ldots$ are $\Sp(n)$ indices
associated with $V$ and running from $1$ to $2n$.  The bundle $\Sigma$
being trivial allows a constant $\Sp(1)$-invariant symplectic form
$\epsilon_{AB}$; whereas $V$ admits an $\Sp(n)$-invariant symplectic
form $\omega_{ab}$.  In terms of these symplectic forms, the metric
$g_{ij}$ on $T$ can be written as:
\begin{equation*}
g_{ij} \gamma^i_{A a} \gamma^j_{B b} = \epsilon_{AB}
\omega_{ab}~.
\end{equation*}
Because the holonomy lies in $\Sp(n)$, not just the metric $g$, but
also the symplectic forms $\epsilon$ and $\omega$ are parallel; whence
so are the maps $\gamma^i_{A a}$.  We choose to trivialise $\Sigma$
globally and put on it the zero connection.  This way any constant
section is parallel.

A final piece of notation is to choose an explicit realisation for the
Clifford algebra in $5{+}1$ dimensions (this will determine the
explicit form of the reality condition satisfied by the fermions).
The metric is $\eta_{mn} = {\rm diag}(-1,1,1,1,1,1)$.
We choose the ($5+1$)--dimensional Gamma matrices to be
\begin{equation*}
\Gamma_\mu = \begin{pmatrix}
0 & \gamma_\mu \\
\gamma_\mu & 0
\end{pmatrix}\quad,\quad
\Gamma_4 = \begin{pmatrix}
0 & -\mathbb{1} \\ \mathbb{1} & 0
\end{pmatrix}\quad\text{and}\quad
\Gamma_5 = \begin{pmatrix}
0 & \gamma_5 \\ \gamma_5 & 0
\end{pmatrix}~,
\end{equation*}
where $\gamma_\mu$, for $\mu=0,1,2,3$, are the ($3{+}1$)-dimensional
gamma matrices:
\begin{equation*}
\gamma^{\mu} = \begin{pmatrix} 0 & \sigma^\mu \\ \Bar\sigma^\mu & 0
\end{pmatrix}\qquad\text{and}\qquad \gamma_5 =
\gamma_0\gamma_1\gamma_2\gamma_3~.
\end{equation*}
Finally, we choose $\Gamma_7 =
\Gamma_0\Gamma_1\Gamma_2\Gamma_3\Gamma_4\Gamma_5$.

We can now write down the following lagrangian (see, e.g.,
\cite{RozanskyWitten}):
\begin{equation*}
\EuScript{L} = \tfrac{1}{2} g_{ij} \d_m \phi^i \d^m \phi^j + \tfrac{1}{2}
\omega_{ab} \Bar\Psi^a \Gamma^m D_m \Psi^b - \tfrac{1}{48}
\Omega_{abcd} (\Bar\Psi^a \Gamma_m \Psi^b) (\Bar\Psi^c
\Gamma^m\Psi^d)~.
\end{equation*}

In this expression,
\begin{itemize}
\item the $\Psi^a$ are eight-component positive-chirality Weyl
spinors: $\Gamma_{7} \Psi^a = + \Psi^a$.  In the above basis for the
$\Gamma$-matrices, it means that $\Psi^a =
\left(\begin{smallmatrix}\psi^a \\ 0 \end{smallmatrix}\right)$, with
$\psi^a$ a four-component complex Dirac spinor, $\psi^a =
\left(\begin{smallmatrix} \chi^a_{\alpha} \\
\Bar{\phi}^{a\,\Dot{\alpha}}\end{smallmatrix}\right)$.
\item the $\Psi^a$ {\it also} satisfy the following symplectic
Majorana condition:
\begin{equation*}
\Psi_a^* \equiv \left(\Psi^a\right)^* = \omega_{ab} \mathsf{B} \Psi^b~,
\end{equation*}
where the matrix $\mathsf{B}$ must satisfy \cite{KugoTownsend} $\Gamma_m^* =
\mathsf{B}\Gamma_m \mathsf{B}^{-1}$, $\mathsf{B}^t=-\mathsf{B}$ and $\mathsf{B}^\dagger \mathsf{B}= \mathbb{1}$.  In our
choice of basis we can take $\mathsf{B}=\Gamma_2\Gamma_5$.  This condition
relates further the two-component spinors comprising $\psi^a$ in such
a way that $\psi^a = \left(\begin{smallmatrix} \chi^a_{\alpha} \\ -
\omega^{ab} \epsilon^{\Dot\alpha\Dot\beta}
\Bar\chi_{b\Dot\beta}\end{smallmatrix}\right)$, where
$\Bar\chi_{b\Dot\beta} \equiv \left( \chi^b_{\beta} \right)^*$.
\item the conjugate $\Bar\Psi^a$ is given by $\Bar\Psi^a = (\Psi^a)^t
\mathsf{C}$, where $\mathsf{C}$ is the charge conjugation matrix, which in our basis
will be chosen to be $\mathsf{C} = -\Gamma_2\Gamma_5\Gamma_0$;
\item the covariant derivative is given by
\begin{equation*}
D_m \Psi^a = \d_m \Psi^a + \Hat\Gamma_{i~b}^a \d_m\phi^i \Psi^b~,
\end{equation*}
with $\Hat\Gamma$ the reduction to $\Sp(n)$ of the riemannian
connection; and
\item $\Omega_{abcd}$ is the hyperK\"ahler curvature, a totally
symmetric tensor defined by
\begin{equation*}
\gamma^i_{A a} \gamma^j_{B b} \gamma^k_{C c}
\gamma^\ell_{D d} R_{ijk\ell} = \epsilon_{AB}\epsilon_{CD}
\Omega_{abcd}~.
\end{equation*}
\end{itemize}

The above lagrangian is invariant under supersymmetry transformations:
\begin{align*}
\delta_\varepsilon \phi^i &= \gamma^i_{A a} \Bar\varepsilon^A\Psi^a\\
\delta_\varepsilon \Psi^a &= \gamma_i^{A a} \Gamma^m \d_m \phi^i
\varepsilon_A - \Hat\Gamma_{i~b}^a \delta_\varepsilon\phi^i \Psi^b
\end{align*}
where $\varepsilon^A$ is a constant {\em negative\/}-chirality Weyl
spinor with values in $\Sigma$ subject to the symplectic Majorana
condition:
\begin{equation*}
\varepsilon_A^* \equiv \left(\varepsilon^A\right)^* = \epsilon_{AB}
\mathsf{B} \varepsilon^B~,
\end{equation*}
and $\Bar\varepsilon^A = \left(\varepsilon^A\right)^t \mathsf{C}$.  The
Noether current generating the supersymmetry is given by
\begin{equation*}
S^{Am} = \omega_{ab} \gamma_i^{Ab} \d_n\phi^i\Gamma^n\Gamma^m\Psi^a~.
\end{equation*}

The above lagrangian is also invariant under the following
infinitesimal Lorentz transformations satisfying the Lie algebra of
$\mathrm{so}(5,1)$:
\begin{align*}
\delta_\Lambda \phi^i &= - \Lambda_{mn} x^m \d^n \phi^i\\
\delta_\Lambda \Psi^a &= -\Lambda_{mn} x^m \d^n \Psi^a + \tfrac{1}{2}
\Lambda_{mn} \Sigma^{mn}\Psi^a~.
\end{align*}
As in the four-dimensional sigma model, we compute the Noether
current by letting $\Lambda_{mn}$ depend on the position and varying
the lagrangian. The time (zeroth) component of this current, whose
integral over space yields the conserved quantities, is found to be
\begin{equation}\label{eq:hkic}
J_{mn} = \tfrac{1}{4} \omega_{ab} \Bar\Psi^a\Gamma^0
\Gamma_{mn}\Psi^b,~
\end{equation}
where $\Gamma_{mn} = \tfrac{1}{2}(\Gamma_m\Gamma_n -
\Gamma_n\Gamma_m)$, and where, in anticipation, we have omitted terms
which will not survive the dimensional reduction.

We now retain $x^0$ as the time and drop all dependence on the other
coordinates.  It is then a simple matter to impose the symplectic
Majorana Weyl condition upon the fermions and use the explicit
realisation for the $\Gamma$-matrices, to derive dimensionally reduced
expressions for the quantities of interest.  We will do this later for
the `internal' symmetry generators.

Finally, one finds after a little calculation that the term in the
Lagrangian density which is quadratic in the fermions may be written
$i\bar\chi_a \Bar\sigma^0\Dot\chi^a$.  After quantisation, one thus
finds the anticommutation relations amongst the fermions
\begin{equation*}
\{\chi^a_\alpha, \Bar\chi_{b\Dot\alpha} \} = - \delta^a_b
\sigma^0_{\alpha\Dot\alpha}~,
\end{equation*}
where again we remind the reader that in our conventions, $\sigma^0 =
-\mathbb{1}$.
We will choose the $\chi^a_\alpha$ as creation operators and the
$\bar\chi_{a\Dot\alpha}$ as annihilation operators, acting on the
appropriate Clifford vacuum.  The bosons are quantised in the usual
way, their Hilbert space being the space of square-integrable
functions $f(\phi)$.  Just as in the K\"ahler case, the total Hilbert
space is (the completion of) the space of smooth complex-valued
differential forms $A_c(X)$ on the hyperK\"ahler manifold $X$.  The
explicit map is the following:
\begin{equation}\label{eq:geoisohk}
f_{i_1i_2\cdots i_k}(x) dx^{i_1}\wedge dx^{i_2}\wedge \cdots\wedge
dx^{i_k} \leftrightarrow f_{i_1i_2\cdots i_k}(\phi) \chi^{i_1}
\chi^{i_2} \cdots \chi^{i_k}|0\rangle~,
\end{equation}
where
\begin{equation*
\chi^i \equiv \gamma^i_{a1} \chi^{a1} + \gamma^i_{a2} \Bar\chi^{a2}~.
\end{equation*}
Notice that in this equation we have identified the index $A$
pertaining to $\Sigma$ with the index $\alpha$.  The inner product is
defined as follows: if $|\psi\rangle = \chi^{a_1}_{\alpha_1}
\chi^{a_2}_{\alpha_2} \cdots \chi^{a_k}_{\alpha_k} |0\rangle$, its
norm is given by $\langle\psi|\psi\rangle$ where
\begin{equation}\label{eq:hknorm}
\langle\psi| = \langle0|
\Bar\chi_{a_k\Dot\alpha_k} \Bar\chi_{a_{k-1}\Dot\alpha_{k-1}}
\cdots\Bar\chi_{a_1\Dot\alpha_1}~.
\end{equation}
In other words, just as in the K\"ahler case the adjoint is complex
conjugation: $\left(\chi^a_\alpha\right)^\dagger =
\Bar\chi_{a\Dot\alpha} = \left(\chi^a_\alpha\right)^*$.  We see that
again the Hilbert space of the quantum mechanical sigma model is (the
completion of) the space of complex-valued smooth differential forms
relative to the Hodge inner product.  It is now possible to quantise
the supercharges and compute their algebra. As is well-known, one
finds that the hamiltonian agrees (up to a factor) with the Hodge
laplacian.  Therefore the ground states are once again in one-to-one
correspondence with the cohomology.

As in the K\"ahler case, it is possible to make a dictionary relating
the quantities appearing in the sigma model with the geometry of $X$.
However we will simply mention that again the supercharges go to
differential operators.  Choosing a complex structure on $X$, it is
possible to single out a particular linear combination of the
supercharges which can be identified with the Dolbeault operator.  In
fact, on $X$ there is a 2-sphere's worth of complex structures, on
which the Lie group $\mathrm{SO}(3)$ acts transitively.  On the sigma model
side, this is nothing but the R-symmetry of the six-dimensional
supersymmetry algebra.  There are also analogues of the Hodge
identities.  Just as in the K\"ahler case, these identities reflect
the transformation properties of the supercharges under the `internal'
symmetry subgroup $\mathrm{Spin}^0(4,1) \cong \Sp(1,1)$ of $\mathrm{Spin}^0(5,1) \cong
\mathrm{SL}(2,\mathbb{H})$.  The supercharges, being Weyl spinors, transform as the
canonical irreducible representation of $\mathrm{SL}(2,\mathbb{H})$ of quaternionic
dimension 2, which under the `internal' $\Sp(1,1)$ subgroup remains
irreducible.

We now come to the main point of this section: the
supersymmetric origins of the action of $\mathrm{so}(4,1)$ on $H^*(X)$.
As before, this action will come induced from the action of the
Lorentz generators under the reduction to one dimension. The
generators $J_{mn}$, for $m,n = 1,...,5$, now generate an `internal'
$\mathrm{so}(5)$ symmetry of the one-dimensional supersymmetric quantum model
obtained by this reduction.  By virtue of its six-dimensional origin,
this symmetry will furthermore commute with the action of the
Hamiltonian, hence will induce an action on the ground states, i.e.,
on the cohomology of $X$.  It remains to show that out of this
$\mathrm{so}(5)$ symmetry follows the $\mathrm{so}(4,1)$ symmetry introduced by
Verbitsky and  summarised in the previous section.  We do so now.

It is enough to consider the linear combinations $L_{i} = -(J_{i5} + i
J_{i4})$.  Expanding the expression \eqref{eq:hkic}, one finds that
\begin{align}
L_i &= -\tfrac12 \omega_{ab} (\chi^a)^t \sigma^2\sigma^i\chi^b\notag\\
\intertext{or more explicitly}
L_1 &= \phantom{-}\tfrac{i}{2} \omega_{ab} \left( \chi^a_1\chi^b_1 -
 \chi^a_2\chi^b_2 \right)\notag\\
L_2 &= -\tfrac12 \omega_{ab} \left( \chi^a_1\chi^b_1 +
 \chi^a_2\chi^b_2 \right)\notag\\
L_3 &= -\tfrac{i}{2} \omega_{ab} \left( \chi_1^a \chi_2^b + \chi_2^a
\chi_1^b \right)~.\label{eq:Liexplicit}
\end{align}
We define $\Lambda_i$ as the adjoints of $L_i$ with respect to the
inner product defined by the norm \eqref{eq:hknorm}, and then the
other operators are defined using the Lie algebra of $\mathrm{so}(4,1)$. In
other words, $K_i$ and $H$ are defined by $K_i = \tfrac12
\epsilon_{ijk}[L_j,\Lambda_k]$, and $H=[L_1,\Lambda_1]$, say.  One can
then check that these operators satisfy the Lie algebra $\mathrm{so}(4,1)$ as
written in \eqref{eq:so(4,1)}.

To conclude the proof we must show that this $\mathrm{so}(4,1)$ agrees with
the one introduced by Verbitsky.  Clearly it is sufficient to check
that the $L_i$ can be interpreted as exterior product with the three
K\"ahler forms.  Under the isomorphism \eqref{eq:geoisohk}, $L_i$
indeed corresponds to exterior product with a 2-form $\omega_i \in
\bigwedge^2T^* \cong \bigwedge^2(\Sigma\otimes V)$ with components
$\omega^i_{aAbB} = \omega_{ab} M^i_{AB}$, where $(M^i)^{AB}$ are
matrices which can be read off from \eqref{eq:Liexplicit}:
\begin{equation*}
M^1 = \begin{pmatrix}i & 0 \\ 0 & -i\end{pmatrix}\quad
M^2 = \begin{pmatrix}-1 & 0 \\ 0 & -1\end{pmatrix}\quad
M^3 = \begin{pmatrix}0 & -i \\ -i & 0\end{pmatrix}~.
\end{equation*}
It is clear that the forms $\omega^i$ are parallel, since so are
$\omega$ and any constant section of any power of $\Sigma$ (recall
that $\Sigma$ is a trivial bundle with the zero connection).  It is
moreover clear that these forms are linearly independent.  Since in an
otherwise arbitrary compact hyperk\"ahler manifold $X$ the space of
parallel 2-forms is three-dimensional and spanned by the K\"ahler
forms, we conclude that these are the K\"ahler
forms.  However we can also
explicitly write down the complex structures in this basis.  The
complex structures are defined by $(I^i)^{aA}_{bB} =  \omega^i_{cCbB}
= \delta^a_b (J^i)^A_B$, where the matrices $J^i$ are related to the
$M^i$ by $(J^i)^A_B = (M^i)^{AC} \epsilon_{CB}$:
\begin{equation*}
J^1 = \begin{pmatrix}0 & i \\ i & 0 \end{pmatrix}\quad
J^2 = \begin{pmatrix}0 & -1\\ 1 & 0 \end{pmatrix}\quad
J^3 = \begin{pmatrix}i & 0 \\  0 & -i\end{pmatrix}~.
\end{equation*}
It is easy to see that they obey the algebra of the imaginary
quaternions:
\begin{equation*}
J^i\,J^j = - \delta^{ij} \mathbb{1} + \epsilon^{ijk} J^k~,
\end{equation*}
as expected.

\section{Conclusions}

In this paper we have shown, following an idea of Witten, how the
Hodge--Lefschetz theory for (hyper)K\"ahler manifolds is a natural
consequence of supersymmetry.  That such a statement can be made
should not come as a surprise because K\"ahler and hyperK\"ahler
manifolds can be {\em defined\/} by the fact that it is
only on these manifolds that the
($3{+}1$)- and ($5{+}1$)-dimensional sigma models admit
supersymmetry.  That this is the case is also linked in a special way
to the properties of Clifford algebras in those dimensions: Weyl
spinors in ($3{+}1$) dimensions are complex, whereas in ($5{+}1$)
dimensions they are quaternionic.  Furthermore, that the
dimensionally reduced sigma model should have anything to do with the
Hodge--Lefschetz theory (or Verbitsky's extension) is again not
surprising, since this theory deals with the cohomology of the
manifold, which are the quantum-mechanical ground states of the
one-dimensional sigma model.

It is tempting to speculate that other aspects of the geometry of
these manifolds also have supersymmetric origins.  For example, if $X$
is compact hyperK\"ahler, Verbitsky \cite{Verbitsky} proved that in
general, there is an action of $\mathrm{so}(4,b_2-2)$ on $H^*(X)$, where $b_2
= \dim H^2(X)$.  It remains an open problem to find a supersymmetric
origin to this symmetry.

\begin{ack}
We would like to acknowledge useful conversations with Jerome
Gauntlett and Chris Hull.  In addition, one of us (JMF) would like to
thank the Isaac Newton Institute for Mathematical Sciences of the
University of Cambridge for its hospitality and support during the
early and final stages of this work.
\end{ack}