We start with a $7$-dimensional complex vector space $V$ and write $Gr_{3}(V)$ for the Grassmann manifold of 3-dimensional subspaces of $V$. So $Gr_{3}(V)$ has dimension $3.(7-3)=12$. A form $\Omega\in\Lambda^{2}(V^{*})$ defines a subset $Z_{\Omega}\subset Gr_{3}(V)$ consisting of the $3$-planes $P$ such that $\Omega|_{P}$ vanishes. In other language we consider the tautological rank 3 vector bundle $U\rightarrow Gr_{3}(V)$; the form $\Omega$ defines a section $s_{\Omega}$ of $\Lambda^{2}U^{*}$ with zero set $Z_{\Omega}$. For generic $\Omega$ this zero set is a smooth subvariety of codimension $3$. Now let $\Omega_{1},\Omega_{2},\Omega_{3}$ be three such forms and consider

$$X=Z_{\Omega_{1}}\cap Z_{\Omega_{2}}\cap Z_{\Omega_{3}}\subset Gr_{3}(V).$$

Of course this only depends on the $3$-plane $\Pi$ in $\Lambda^{2}V^{*}$ spanned by the $\Omega_{i}$, so we may sometimes write $X_{\Pi}$. Obviously there is a Zariski-open subset ${\cal U}$ in the Grassmannian $Gr_{3}(\Lambda^{2}V^{*})$ of $3$-planes $\Pi$ such that $X_{\Pi}$ is a smooth subvariety of dimension $12-3.3=3$. This set ${\cal U}$ is non-empty, as we will see later. The group $SL(V)$ acts on the whole construction and obviously different subspaces $\Pi$ which lie in the same $SL(V)$ orbit define isomorphic manifolds $X_{\Pi}$, so we get a set of equivalence classes of manifolds constructed in this way, parametrised by the quotient ${\cal U}/SL(V)$. (Mukai shows further that this parametrisation is effective: i.e. $X_{\Pi_{1}}$ is isomorphic to $X_{\Pi_{2}}$ if and only if $\Pi_{1},\Pi_{2}$ lie in the same $SL(V)$ orbit. Moreover, he shows that all “prime Fano $3$-folds of genus 12” arise in this way.)

We compute the canonical bundle $K_{X}$ of the variety $X=X_{\Pi}$ for some $\Pi\in{\cal U}$. We have

$$\Lambda^{2}U^{*}=U\otimes H$$

where $H$ is the ample line bundle $\Lambda^{3}U^{*}$. So, writing $\det$ for the the top exterior power of a vector bundle, we have

$$\det\Lambda^{2}U^{*}=H^{\otimes 2}.$$

The tangent bundle of the Grassmannian at a $3$-plane $P\subset V$ can be identified with $P^{*}\otimes V/P$. So

$$\det TGr_{3}=H^{\otimes 7}.$$

Now since the tangent bundle of $X$ is the kernel of a surjective map from $TGr_{3}(V)$ to $\Lambda^{2}U^{*}\oplus\Lambda^{2}U^{*}\oplus\Lambda^{2}U^{*}$ we have

$$K_{X}^{-1}=\det TX=H^{\otimes(7-3.2)}=H.$$

Thus $X$ is a Fano manifold. The sections of $H$ over $Gr_{3}$ give the Plucker embedding

$$Gr_{3}(V)\rightarrow{\bf P}(\Lambda^{3}V)={\bf P}^{34}$$

For any $3$-form $A\in\Lambda^{3}V^{*}$ we get a hyperplane section $Y_{A}\subset Gr_{3}(V)$ which just consists of the $3$-planes $P$ such $A|_{P}=0$. By definition this occurs if $P$ is in $X$ and $A$ is in the image of the wedge product map $\Pi\otimes V^{*}\rightarrow\Lambda^{3}V^{*}$. We expect this map to have an image of dimension $7.3=21$ in which case the image of the composite

$$X\rightarrow Gr_{3}(V)\rightarrow{\bf P}^{34}$$

lies in a linear subspace ${\bf P}^{34-21}={\bf P}^{13}$. Certainly this map is defined by sections of $K_{X}^{-1}$, we will see later that $H^{0}(X,K_{X}^{-1})$ has dimension $14$ and that this embedding is that given by the anticanonical system.

To make this more concrete we show now that $X$ is a rational variety; that is, we construct an explicit parametrisation of a dense open set in $X$. Suppose we have a pair of $3$-dimensional subspaces $P_{0},Q_{0}\subset V$ with $P_{0}\cap Q_{0}=0$. We ask what $3$-planes $P$ in the $6$-dimensional subspace $P_{0}\oplus Q_{0}$ lie in $X$. In matrix notation, we can write the restriction of a form $\Omega$ to $P_{0}\oplus Q_{0}$ as

$$\left(\begin{array}[]{cc}\sigma&A\\ -A^{T}&\tau\end{array}\right)$$

Now consider the $3$-dimensional subspaces $P$ which arise as the graphs of linear maps $M:P_{0}\rightarrow Q_{0}$. The condition becomes

$$\sigma+M^{T}\tau M+(AM-(AM)^{T})=0.$$

(33)

So our three forms $\Omega_{i}$ give us three triples $A_{i},\sigma_{i},\tau_{i}$ and we have three equations of the form (33) to solve to find a point of $X$. We have $9$ unknowns: the entries of the matrix $M$. The left hand side of (33) takes values in the $3$-dimensional space of skew symmetric $3\times 3$ matrices so we obtain a total of $3.3=9$ equations in these $9$ unknowns and we expect a finite number of solutions. These equations are quadratic and one can solve them explicitly, to see that there are generically two solutions. However it is easier to suppose that we are in the case when $P_{0}$ itself lies in $X$. This means that all the $\tau_{i}$ are zero, so the equations (33) become linear. Generically this system of $9$ linear equations in $9$ unknowns is nondegerate and there is a unique solution. Now suppose we have found one point $P_{0}$ in $X$ and consider the space of $6$-planes in $V$ which contain $P_{0}$. This is a copy of projective $3$-space ${\bf P}^{3}$. Given a point in ${\bf P}^{3}$, that is to say a 6 dimensional subspace $E$ of $V$, we choose a complementary subspace to write is as $E=P_{0}\oplus Q_{0}$. Then we can proceed as above and, by solving linear equations, find the points of $X\cap Gr_{3}(E)$. Generically there is just one, $P_{E}$ say, different from the original $P_{0}$. Conversely for any $P^{\prime}\in X$ the sum $P\oplus P^{\prime}$ lies in a $6$-dimensional subspace. Of course there will be various exceptional cases, but the upshot is that we get a birational map from ${\bf P}^{3}$ to $X$ which takes a subspace $E$ containing $P_{0}$ to $P_{E}$.

We now consider a special manifold in this family. Take the vector space $V$ to be the sixth symmetric power $s^{6}$ of the fundamental representation of $SL(2,{\bf C})$. Then $\Lambda^{2}V^{*}=\Lambda^{2}s^{6}$ decomposes into distinct irreducible representations

$$\Lambda^{2}s^{6}=s^{10}\oplus s^{6}\oplus s^{2}.$$

The $s^{2}$ summand is a $3$-plane $\Pi_{0}$ invariant under $SL(2,{\bf C})$, so there is a natural $SL(2,{\bf C})$ action on the corresponding variety, the Mukai-Umemura manifold, $X_{0}=X_{\Pi_{0}}$. We will see below that $X_{0}$ admits a Kahler-Einstein metric. The representation $s^{6}$ has a standard invariant symmetric form $(\ ,\ )$ and the inclusion $s^{2}\rightarrow\Lambda^{2}s^{6}$ is just the map from the Lie algebra of $SL(2,{\bf C})$ given by the action on $s^{6}$. This comes down to saying that a $3$-plane $P$ is in $X_{0}$ if and only if

$$(\delta p,q)=0$$

(34)

for all $p,q\in P$ and $\delta\in\mathfrak{s}\mathfrak{l}_{2}$. Notice that the action of $SL(2,{\bf C})$ on all the spaces involved actually factors through $PSL(2,{\bf C})$.

Identify the projectivisation of the fundamental representation $s^{1}={\bf C}^{2}$ with the standard round sphere and fix an icosahedron, which can be regarded as a set of 12 vertices in in this sphere. Thus we get a symmetry group $\Gamma\subset SO(3)\subset PSL(2,{\bf C})$ of order $60$. There is a simple way to see that the $7$-dimensional representation $s^{6}$ of $PSL(2,{\bf C})$ becomes reducible when restricted to $\Gamma$. There are $6$ pairs of antipodal vertices and for each such pair $p,\overline{p}$ we have a $1$-dimensional subspace consisting of polynomials which vanish to order $3$ at $p,\overline{p}$. The sum of these $6$ subspaces is obviously invariant under $\Gamma$ and is a proper subspace of $s^{6}$ since it has codimension at least $1$. A little calculation shows that this invariant subspace is of dimension $3$ and satisfies the criterion (34). So this subspace gives a point $P_{0}$ in $X_{0}$ fixed by $\Gamma$. On the other hand the stabiliser of $P_{0}$ is obviously not the whole of $SO(3)$ and, since there is no finite subgroup of $SO(3)$ strictly larger than $\Gamma$, the stabiliser must be exactly $\Gamma$.

Now go back to the wedge product $P_{0}\wedge V^{*}\rightarrow\Lambda^{3}V^{*}$. In terms of representations this is an $SL(2,{\bf C})$-map

$$s^{2}\otimes s^{6}\rightarrow\Lambda^{3}s^{6}.$$

It is an exercise in representation theory to show that

$$\Lambda^{3}s^{6}=s^{12}\oplus s^{8}\oplus s^{6}\oplus s^{4}\oplus s^{2}\oplus s^{0}\oplus s^{0}.$$

So comparing with

$$s^{2}\otimes s^{6}=s^{8}\oplus s^{6}\oplus s^{4}\oplus s^{2}\oplus s^{0}$$

we see that the embedding $X_{0}\subset Gr_{3}(V)\subset{\bf P}(\Lambda^{3}V)$ gives rise to an $SL(2,{\bf C})$-equivariant embedding

$$X_{0}\rightarrow{\bf P}(s^{0}\oplus s^{12}).$$

(35)

In other words, by our identification of the anticanonical bundle $K^{-1}$ we have

$$H^{0}(X_{0},K^{-1})=s^{0}\oplus s^{12},$$

as a representation of $SL(2,{\bf C})$.In particular, there is an $SL(2,{\bf C})$-invariant section $\sigma$ of $K^{-1}$. Explicitly, if we identify $\Lambda^{3}s^{6}$ with $\Lambda^{4}s^{6}$ then $\sigma$ corresponds to the $4$-form on $V=s^{6}$ defined as follows. We choose any orthonormal basis $\Omega_{1},\Omega_{2},\Omega_{3}$ of $P_{0}$ and write down the $4$-form

$$*\sigma=\Omega_{1}^{2}+\Omega_{2}^{2}+\Omega_{3}^{2}.$$

In this way, we get another description of the manifold $X_{0}$. Our point $P_{0}\in X_{0}$ cannot lie in the zero set of $\sigma$ (since its orbit is $3$-dimensional). So, in the embedding (35), we have

$$P_{0}=[1,v_{0}]\in{\bf P}({\bf C}\oplus s^{12}).$$

Thus $v_{0}$ is an element of $s^{12}$ whose stabiliser in $PSL(2,{\bf C})$ is exactly $\Gamma$.Now there is an obvious element of the projective space ${\bf P}(s^{12})$ with stabiliser $\Gamma$, just the configuration of vertices of the icosahedron, regarded as an element of the symmetric product. Since $\Gamma$ is a perfect group it must act trivially on the corresponding line in $s^{12}$, so we get a vector in $s^{12}$ with stabiliser $\Gamma$. It is easy to see that, up to a multiple, this in the only element of $s^{12}$ with stabiliser $\Gamma$, and thus we have identified $v_{0}$. Then we can simply define $X_{0}$ to be the closure in ${\bf P}({\bf C}\oplus s^{12})$ of the $PSL(2,{\bf C})$-orbit of $v_{0}$ in $s^{12}$. (Here we are regarding the vector space $s^{12}$ as being a subset of the projective space ${\bf P}({\bf C}\oplus s^{12})$ in the familiar way.)

In this description, the intersection of $X_{0}$ with the hyperplane at infinity

$$D={\bf P}(s^{12})\subset{\bf P}({\bf C}\oplus s^{12}),$$

is, by definition, the zero set of the invariant section $\sigma$ of $K^{-1}$. Consider a $1$-parameter subgroup $\lambda_{t}$ in $PSL(2,{\bf C})$. Thus we have a pair of distinct point $z_{+},z_{-}$ such that when $t$ is large positive the map $\lambda_{t}$ contracts most of the sphere to a small neighbourhood of $z_{+}$, and when $t$ is large negative to a small neighbourhood of $z_{-}$. If $y_{1},\dots y_{12}$ is any configuration of distinct points it is not hard to see that the limit as $t\rightarrow\infty$ of

$$\lambda_{t}(\underline{y})=\left(\lambda_{t}(y_{1}),\lambda_{t}(y_{2}\dots\lambda_{t}(y_{12})\right)$$

in the symmetric product ${\bf P}(s^{12})$ is either $12z_{+}=(z_{+},z_{+},\dots,z_{+})$ (in the generic case) or $11z_{+}+z_{-}=(z_{+},\dots,z_{+},z_{-})$ (in the case when one of the $y_{i}$ is $z_{-}$). Using this, Mukai and Umemura show that the divisor at infinity $D$ consists precisely of the union of points of the form $12z_{+}$ or $11z_{+}+z_{-}$ in ${\bf P}(s^{12})$. It is easy to identify this geometrically. The points of the form $12z_{+}$ make up the rational normal curve in ${\bf P}(s^{12})$. Our divisor $D$ is the surface swept out by the lines in ${\bf P}(s^{12})$ tangent to the rational normal curve. As a set we can identify $D$ with ${\bf P}^{1}\times{\bf P}^{1}$: we just map $(z_{+},z_{-})\in{\bf P}^{1}\times{\bf P}^{1}$ to $11z_{+}+z_{-}\in D$. But the surface $D$ is singular and a more precise statement is that the map above is a holomorphic map $\nu:{\bf P}^{1}\times{\bf P}^{1}\rightarrow D$ which is the normalisation of $D$. The singular set of $D$ is the image of the diagonal in ${\bf P}^{1}\times{\bf P}^{1}$, and it is easy to check that the singularity has the form of a cusp transverse to the diagonal. That is to say, we can choose local co-ordinates $z_{1}z_{2},z_{3}$ in $X_{0}$ around a singular point of $D$ such that $D$ is defined by the equation $z_{1}^{2}=z_{2}^{3}$.

We now have a rather explicit description of $X_{0}$, as the compactification of $PSL(2,{\bf C})/\Gamma$ formed by adjoining the divisor $D$. We can use this to compute the action of $PSL(2,{\bf C})$ on all of the spaces of sections $H^{0}(X_{0},K^{-p})$. For the pull back $\nu^{*}(K^{-1})$ is isomorphic to the line bundle ${\cal O}(11,1)$ over ${\bf P}^{1}\times{\bf P}^{1}$. We can regard the structure sheaf of $D$ as a subsheaf of that of ${\bf P}^{1}\times{\bf P}^{1}$. From the local model of the singularity along the diagonal one sees that the quotient can be identified with sections of ${\cal O}(2)$ along the diagonal. This means that $H^{0}(D,K^{-p}|_{D})$ is the kernel of a map $H^{0}({\bf P}^{1}\times{\bf P}^{1};{\cal O}(11p,p))\rightarrow H^{0}({\bf P}^{1};{\cal O}(12p-2))$. As representations of $PSL(2,{\bf C})$ this is a map

$$s^{11p}\otimes s^{p}\rightarrow s^{12p-2}.$$

Now

$$s^{11p}\otimes s^{p}=s^{12p}+s^{12p-2}\dots\oplus s^{10p}$$

and the map above is just the projection to the second factor. So

$$H^{0}(D;K^{-p}|_{D})=s^{12p}\oplus s^{12p-4}\oplus s^{12p-6}\dots\oplus s^{10p+2}\oplus s^{10p}.$$

Then the exact cohomology sequence of

$$0\rightarrow K^{-(p-1)}\rightarrow K^{-p}\rightarrow K^{-p}|_{D}\rightarrow 0$$

together with Kodaira vanishing on $X_{0}$ gives

$$H^{0}(X_{0},K^{-p})=H^{0}(X_{0},K^{-(p-1)})\oplus s^{12p}\oplus s^{12p-4}\dots s^{10p},$$

and inductively we get a description of each $H^{0}(X_{0},K^{-p})$. Thus

$$H^{0}(X_{0},K^{-1})=s^{0}\oplus s^{12},$$

$$H^{0}(X_{0},K^{-2})=s^{0}\oplus s^{12}\oplus s^{24}\oplus s^{20}.$$

For $p\geq 6$ we get multiplicities: $H^{0}(X_{0},K^{-6})$ contains two copies of $s^{60}$. This illustrates the difference with the multiplicity-free case discussed above. (Although since the multiplicities are small until $p$ becomes quite large, once is tempted to think of $X_{0}$ as being “close” to multiplicity-free. )