# Solution to exercises

Exercise 8.3

Let ${\displaystyle V}$ be a real vector space with ${\displaystyle \dim _{\mathbb {R} }=2n}$. For ${\displaystyle \omega \in \bigwedge ^{2}V^{*}}$ show that ${\displaystyle \omega ^{\wedge n}\neq 0}$ if and only if the map ${\displaystyle \lrcorner \omega :V\rightarrow V^{*}}$ is an isomorphism.

Proof

Let's start near the beginning and describe the map ${\displaystyle \lrcorner \omega }$. Recall the identification of ${\displaystyle V^{*}\otimes V^{*}}$ with the collection of bilinear forms on ${\displaystyle V}$: if ${\displaystyle \alpha ,\beta \in V^{*}}$ then ${\displaystyle \alpha \otimes \beta }$ corresponds to the bilinear form ${\displaystyle (v,w)\mapsto \alpha (v)\beta (w)}$. Now every bilinear form can be expressed as a sum of a symmetric and an anti-symmetric bilinear form:

${\displaystyle \alpha (v,w)={\frac {\alpha (v,w)+\alpha (w,v)}{2}}+{\frac {\alpha (v,w)-\alpha (w,v)}{2}}.}$
On the ${\displaystyle V^{*}\otimes V^{*}}$ side of this picture this fact corresponds to the usual decomposition ${\displaystyle S^{2}V^{*}\oplus \bigwedge ^{2}V^{*}}$ of ${\displaystyle V^{*}\otimes V^{*}}$. As such we get an identification between ${\displaystyle \bigwedge ^{2}V^{*}}$ and the collection of anti-symmetric bilinear forms on ${\displaystyle V}$. Explicitly
${\displaystyle \alpha \wedge \beta \mapsto {\big (}(v,w)\mapsto {\frac {\alpha (v)\beta (w)-\alpha (w)\beta (v)}{2}}{\big )}.}$
In view of the above we may think of ${\displaystyle \omega }$ as an anti-symmetric bilinear form on ${\displaystyle V}$; the map ${\displaystyle \lrcorner \omega }$ sends a vector ${\displaystyle v\in V}$ onto ${\displaystyle w\mapsto \omega (v,w)}$, an element of ${\displaystyle V^{*}}$.

With these identifications fixed, choosing a basis for ${\displaystyle V}$ allows us to represent our ${\displaystyle \omega }$ as an anti-symmetric square matrix and in fact, so long as we choose our basis carefully, this matrix may be taken to be of the following form:

${\displaystyle {\begin{pmatrix}0&\lambda _{1}&&&&&&&&\\-\lambda _{1}&0&&&&&&&&\\&&0&\lambda _{2}&&&&&&\\&&-\lambda _{2}&0&&&&&&\\&&&&\ddots &&&&&\\&&&&&0&\lambda _{r}&&&\\&&&&&-\lambda _{r}&0&&&\\&&&&&&&0&&\\&&&&&&&&\ddots &\\&&&&&&&&&0\\\end{pmatrix}}}$
If we denote this basis by ${\displaystyle \lbrace e_{i}\rbrace }$ and let ${\displaystyle \lbrace e_{i}^{*}\rbrace }$ be the dual basis of ${\displaystyle V^{*}}$ then the fact that the matrix of ${\displaystyle \omega }$ has such a nice description corresponds to the fact that we can write ${\displaystyle \omega =\sum _{i=1}^{n}\lambda _{i}(e_{2i-1}^{*}\wedge e_{2i}^{*})}$ in ${\displaystyle \bigwedge ^{2}V^{*}}$ (where ${\displaystyle \lambda _{i}=0}$ for ${\displaystyle i>r}$). To see that this is the case we look at the ${\displaystyle \lambda _{i}(e_{2i-1}^{*}\wedge e_{2i}^{*})}$ term. In the world of bilinear forms on ${\displaystyle V}$ this looks like
${\displaystyle (v,w)\mapsto \lambda _{i}{\frac {e_{2i-1}^{*}(v)e_{2i}^{*}(w)-e_{2i-1}^{*}(w)e_{2i}^{*}(v)}{2}}}$
and on the subspace of ${\displaystyle V}$ spanned by ${\displaystyle e_{2i-1}}$ and ${\displaystyle e_{2i}}$ this acts just as ${\displaystyle \omega }$ did.

The scene is set for us to ask: when is ${\displaystyle \omega ^{\wedge n}}$ non-zero?. As the wedge product distributes over addition we can expand the brackets and view ${\displaystyle \omega ^{\wedge n}}$ as a sum

${\displaystyle \sum (n{\text{-fold products of the }}\lambda _{i}{\text{s}})(n{\text{-fold wedge products of the }}e_{2i-1}^{*}\wedge e_{2i}^{*}{\text{s}}).}$
Note that an ${\displaystyle n}$-fold wedge product of the ${\displaystyle e_{2i-1}^{*}\wedge e_{2i}^{*}}$s is non-zero precisely when all the ${\displaystyle e_{i}}$s appear in this wedge product; there is only one such summand in the above and we can choose the ordering in ${\displaystyle n!}$ ways. As we are commuting 2-forms, rearranging the terms into ascending order doesn't introduce any more signs and so ${\displaystyle \omega ^{\wedge n}}$ is equal to
${\displaystyle n!(\prod _{i=1}^{n}\lambda _{i})(e_{1}^{*}\wedge \ldots \wedge e_{2n}^{*}).}$
Evidently this is non-zero if and only if all the ${\displaystyle \lambda _{i}}$s are all non-zero, i.e., if and only if ${\displaystyle r=n}$, i.e., if and only if the matrix representing ${\displaystyle \omega }$ is invertible. This being the precise condition which describes when the map ${\displaystyle \lrcorner \omega }$ is an isomorphism we find that the exercise is complete.

Exercise 8.4

Let ${\displaystyle O(n)}$ act on ${\displaystyle \mathbb {R} ^{n}}$ in the standard way and consider the induced action on the cotangent space ${\displaystyle T^{*}\mathbb {R} ^{n}}$. Show that the moment map is angular momentum.

Proof

Let ${\displaystyle (x_{1},\ldots ,x_{n})}$ be coordinates on ${\displaystyle \mathbb {R} ^{n}}$ and ${\displaystyle (x_{1},\ldots ,x_{n};\xi _{1},\ldots ,\xi _{n})}$ coordinates on ${\displaystyle M=T^{*}\mathbb {R} ^{n}}$.

Then we have natural coordinates ${\displaystyle ({\underline {x}},{\underline {\xi }},{\frac {\partial }{\partial x_{1}}},\ldots ,{\frac {\partial }{\partial x_{n}}},{\frac {\partial }{\partial \xi _{1}}},\ldots ,{\frac {\partial }{\partial \xi _{n}}})}$ on ${\displaystyle TM}$ and ${\displaystyle ({\underline {x}},{\underline {\xi }},dx_{1},\ldots ,dx_{n},d\xi _{1},\ldots ,d\xi _{n})}$ on ${\displaystyle T^{*}M}$.

Recall that the standard symplectic form on ${\displaystyle M}$ is ${\displaystyle \omega =\sum _{i=1}^{n}dx_{i}\wedge d\xi _{i}}$. Since the action on ${\displaystyle \mathbb {R} ^{n}}$ is linear, the action on the cotangent bundle is simply given by ${\displaystyle A.({\underline {x}},{\underline {\xi }})=(A{\underline {x}},{\underline {\xi }}A^{-1})}$ for any ${\displaystyle A\in O(n)=\{Y\in GL_{n}(\mathbb {R} )|\ YY^{t}=I\}}$. It is easy to check that this action is symplectic: in fact, write ${\displaystyle B}$ for ${\displaystyle A^{t}=A^{-1}}$ and observe that the symplectic form goes to ${\displaystyle \sum _{i=1}^{n}(\sum _{j=1}^{n}A_{i}^{j}dx_{j})\wedge (\sum _{k=1}^{n}B_{k}^{i}d\xi _{k})=\sum _{j,k}(\sum _{i}A_{i}^{j}B_{k}^{i})dx_{j}\wedge d\xi _{k}=\sum _{j,k}\delta _{jk}dx_{j}\wedge d\xi _{k}}$.

Recall that the Lie algebra of ${\displaystyle O(n)}$ is ${\displaystyle {\mathfrak {o}}(n)=\{y\in {\mathfrak {gl}}_{n}(\mathbb {R} )|\ y+y^{t}=0\}}$. We want to translate the action into a map ${\displaystyle {\mathfrak {o}}(n)\rightarrow \Gamma (TM)}$; to the matrix ${\displaystyle y}$ we associate the section ${\displaystyle X_{y}\colon ({\underline {x}},{\underline {\xi }})\mapsto {\frac {d}{dt}}_{|t=0}({\underline {x}},{\underline {\xi }},\operatorname {\underline {x}} ,{\underline {\xi }}\operatorname {)} =({\underline {x}},{\underline {\xi }},(y{\underline {x}})_{i}{\underline {\frac {\partial }{\partial x_{i}}}},-({\underline {\xi }}y)_{i}{\underline {\frac {\partial }{\partial \xi _{i}}}})}$. Now construct the ${\displaystyle d}$-closed form ${\displaystyle \omega (X_{y},\cdot )}$, (recall that ${\displaystyle dx_{i}\wedge d\xi _{i}({\frac {\partial }{\partial \xi _{i}}},\cdot )=-dx_{i}}$) and obtain ${\displaystyle \sum _{i}(\sum _{j}x_{j}y_{i}^{j})d\xi _{i}+(\sum _{j}\xi _{j}y_{j}^{i})dx_{i}}$. Observe that this is exact with primitive ${\displaystyle h_{y}({\underline {x}},{\underline {\xi }})=\sum _{i,j}y_{i}^{j}x_{j}\xi _{i}}$ (up to constants).

Finally, ${\displaystyle {\mathfrak {o}}(n)}$ is a real vector space of dimension ${\displaystyle {\frac {n(n-1)}{2}}}$ and a natural basis of its dual is ${\displaystyle \{e_{i}^{j}\}_{i, defined by ${\displaystyle e_{i}^{j}(y)=y_{i}^{j}}$. Since ${\displaystyle y_{i}^{j}=-y_{j}^{i}}$ for our anti-symmetric matrices, the moment map ${\displaystyle m\colon M\to {\mathfrak {o}}(n)^{*}}$ is given by ${\displaystyle m({\underline {x}},{\underline {\xi }})=\sum _{i which can be identified with ${\displaystyle {\underline {x}}\wedge {\underline {\xi }}}$ under ${\displaystyle e_{j}^{i}=e_{j}\wedge e_{i}}$. Compare with angular momentum in classical mechanics ${\displaystyle \mathbf {l} =\mathbf {x} \wedge m{\dot {\mathbf {x} }}}$.

Exercise 8.5

Let ${\displaystyle S^{1}}$ act on ${\displaystyle \mathbb {C} }$ by multiplication. Show that the associated vector fields are ${\displaystyle X_{\xi }(z)=i\xi z}$ for ${\displaystyle \xi \in \operatorname {Lie} (S^{1})\cong \mathbb {R} }$. Show that these can be written as ${\displaystyle \xi {\frac {\partial }{\partial \theta }}}$ for ${\displaystyle z=re^{i\theta }}$. Show that these are Hamiltonian, where the Hamiltonian is given by ${\displaystyle -\xi {\frac {r^{2}}{2}}+c}$.

Proof

The vector fields associated to a Lie group action ${\displaystyle G\curvearrowright M}$ on a manifold ${\displaystyle M}$ are given by ${\displaystyle X_{\xi }(p)=\left.{\frac {d}{dt}}\right|_{t=0}\operatorname {exp} (t\xi )p}$. In our case we have ${\displaystyle \operatorname {exp} (t\xi )=e^{it\xi }}$ and so we obtain

${\displaystyle X_{\xi }(z)=\left.{\frac {d}{dt}}\right|_{t=0}e^{it\xi }z=i\xi z.}$
In Cartesian coordiantes ${\displaystyle z=x+iy}$ this is given by ${\displaystyle \xi x{\frac {\partial }{\partial y}}-\xi y{\frac {\partial }{\partial x}}}$. To translate to polar coordiantes we note that ${\displaystyle x=r\cos \theta }$ and ${\displaystyle y=r\sin \theta }$ and we have to compute
${\displaystyle {\begin{pmatrix}{\frac {\partial x}{\partial r}}&{\frac {\partial x}{\partial \theta }}\\{\frac {\partial y}{\partial r}}&{\frac {\partial y}{\partial \theta }}\end{pmatrix}}^{-1}={\begin{pmatrix}\cos \theta &-r\sin \theta \\\sin \theta &r\cos \theta \end{pmatrix}}^{-1}={\begin{pmatrix}\cos \theta &\sin \theta \\-{\frac {1}{r}}\sin \theta &{\frac {1}{r}}\cos \theta \end{pmatrix}}.}$
From this we read off that ${\displaystyle {\frac {\partial }{\partial x}}=\cos \theta {\frac {\partial }{\partial r}}-{\frac {1}{r}}\sin \theta {\frac {\partial }{\partial \theta }}}$ and ${\displaystyle {\frac {\partial }{\partial y}}=\sin \theta {\frac {\partial }{\partial r}}+{\frac {1}{r}}\cos \theta {\frac {\partial }{\partial \theta }}}$. Substituting this into the Cartesian expression for ${\displaystyle X_{\xi }}$ we obtain ${\displaystyle X_{\xi }=\xi {\frac {\partial }{\partial \theta }}}$ as expected. To compute the Hamiltonian for this vector field we first have to express the symplectic form in polar coordinates. The standard symplectic form on ${\displaystyle \mathbb {C} }$ in Cartesian coordinates is ${\displaystyle \omega =\mathop {} \!dx\wedge \mathop {} \!dy}$. Similarly as before we first compute
{\displaystyle {\begin{aligned}\mathop {} \!dx&=\cos \theta \mathop {} \!dr-r\sin \theta \mathop {} \!d\theta \\\mathop {} \!dy&=\sin \theta \mathop {} \!dr+r\cos \theta \mathop {} \!d\theta ,\end{aligned}}}
and by substitution we obtain ${\displaystyle \omega =\mathop {} \!dx\wedge \mathop {} \!dy=r\mathop {} \!dr\wedge \mathop {} \!d\theta }$. Next we compute the contraction ${\displaystyle X_{\xi }\lrcorner \omega }$:
${\displaystyle X_{\xi }\lrcorner \omega =\omega (X_{\xi },\cdot )=(r\mathop {} \!dr\wedge \mathop {} \!d\theta )\left(\xi {\frac {\partial }{\partial \theta }},\cdot \right)=-\xi r\mathop {} \!dr.}$
Finally we integrate ${\displaystyle X_{\xi }\lrcorner \omega }$ to obtain the Hamiltonian:
${\displaystyle \int -\xi r\mathop {} \!dr=-\xi {\frac {r^{2}}{2}}+c.}$

Exercise 8.6

Let ${\displaystyle X}$ be a symplectic vector field on a compact symplectic manifold ${\displaystyle (M,\omega )}$ and ${\displaystyle \gamma }$ a path in ${\displaystyle M}$. Denote by ${\displaystyle T_{\epsilon }}$ the surface swept out after time ${\displaystyle \epsilon }$ by the flow along the vector field ${\displaystyle X}$ starting at ${\displaystyle \gamma }$. Then define ${\displaystyle {\text{Flux}}_{\gamma }(X):=\lim _{\epsilon \to 0}{\frac {1}{\epsilon }}\int _{T_{\epsilon }}\omega }$. Prove that ${\displaystyle X}$ is Hamiltonian ${\displaystyle \Leftrightarrow }$${\displaystyle {\text{Flux}}_{\gamma }(X)=0}$ for every ${\displaystyle \gamma \in H_{1}(M)}$.

Proof

Since ${\displaystyle X}$ is a symplectic vector field we have that ${\displaystyle X\lrcorner \omega }$ is closed and so represents a cohomology class in ${\displaystyle H^{1}(M)}$. This class is zero (i.e. ${\displaystyle X}$ is a Hamiltonian vector field) precisely when it integrates to zero on every closed loop in ${\displaystyle M}$. So it will suffice to prove that

${\displaystyle Flux_{\gamma }(X)=\int _{\gamma }X\lrcorner \omega .}$
The above is a general fact which holds for any simple path ${\displaystyle \gamma :[0,1]\to M}$, not necessarily closed, and ${\displaystyle \omega }$ and ${\displaystyle X}$ any ${\displaystyle 2-}$form and vector field on ${\displaystyle M}$ respectively. To calculate ${\displaystyle Flux_{\gamma }(X)=\lim _{\epsilon \to 0}{\frac {1}{\epsilon }}\int _{T_{\epsilon }}\omega }$ we can cover ${\displaystyle \gamma ([0,1])}$ by finitely many coordinate neghbourhoods ${\displaystyle \{U_{i}\}_{1\leq i\leq n}}$ each containing ${\displaystyle \gamma ([\alpha _{i},\alpha _{i+1}])}$ for some ${\displaystyle 0=\alpha _{0}<\alpha _{1}<\cdots <\alpha _{n}=1}$. Since the flux only depends on small ${\displaystyle \epsilon }$ we can assume ${\displaystyle T_{\epsilon }\subset \cup _{i}U_{i}}$. We pick a partition of unity ${\displaystyle \{\phi _{ij}\}}$ with ${\displaystyle \phi _{ij}}$ supported in ${\displaystyle U_{i}}$ for every ${\displaystyle j}$ and now by definition ${\displaystyle \int _{T_{\epsilon }}\omega =\Sigma _{i,j}\int _{T_{\epsilon }}\omega \phi _{ij}}$ and ${\displaystyle \int _{\gamma }X\lrcorner \omega =\Sigma _{i,j}\int _{\gamma }X\lrcorner \omega \phi _{ij}}$. So we can prove (Solution to exercises) only locally and the global statement will follow. We will prove the identity in the neighbourhood ${\displaystyle U_{i_{0}}}$ using the function ${\displaystyle \phi _{i_{0}j_{0}}}$. We have the following local expressions (using Einstein summation notation):
${\displaystyle \gamma (s)=(\gamma ^{1}(s),\ldots \gamma ^{n}(s)),}$
${\displaystyle X({\textbf {x}})=a^{i}({\textbf {x}}){\frac {\partial }{\partial {x^{i}}}},}$
${\displaystyle \omega \phi _{i_{0}j_{0}}=\omega _{ij}dx^{i}\wedge dx^{j}.}$
For the flow of ${\displaystyle X}$ we write ${\displaystyle F_{t}({\textbf {x}})=(u^{1}(t,{\textbf {x}}),u^{2}(t,{\textbf {x}}),\ldots ,u^{n}(t,{\textbf {x}}))}$ which satisfies ${\displaystyle {\frac {\partial u^{i}}{\partial {t}}}=a^{i}}$ and ${\displaystyle u^{i}(0,{\textbf {x}})=x^{i}}$. We get a local parametrisation of ${\displaystyle T_{\epsilon }}$ by
${\displaystyle (s,t)\mapsto F_{t}(\gamma (s))=(u^{1}(t,\gamma ^{1}(s),\ldots \gamma ^{n}(s)),\;\ldots ,\;u^{n}(t,\gamma ^{1}(s),\ldots ,\gamma ^{n}(s)))}$
for ${\displaystyle s\in (\alpha _{i_{0}}}$, ${\displaystyle \alpha _{i_{0}+1}),t\in (0,\epsilon )}$. Then we have
${\displaystyle dx^{i}=d\left[u^{i}(t,\gamma ^{1}(s),\ldots \gamma ^{n}(s))\right]={\frac {\partial u^{i}}{\partial t}}dt+{\frac {\partial u^{i}}{\partial x^{k}}}{\frac {d\gamma ^{k}}{ds}}ds}$
${\displaystyle \Rightarrow dx^{i}\wedge dx^{j}=\left({\frac {\partial u^{i}}{\partial t}}{\frac {\partial u^{j}}{\partial x^{k}}}-{\frac {\partial u^{j}}{\partial t}}{\frac {\partial u^{i}}{\partial x^{k}}}\right){\frac {d\gamma ^{k}}{ds}}(dt\wedge ds).}$
So for the flux we have:
{\displaystyle {\begin{aligned}\lim _{\epsilon \to 0}\int _{T_{\epsilon }}\omega \phi _{i_{0}j_{0}}&=&\lim _{\epsilon \to 0}\left({\frac {1}{\epsilon }}\int _{\alpha _{i_{0}}}^{\alpha _{i_{0}+1}}\int _{0}^{\epsilon }\left[\omega _{ij}\left({\frac {\partial u^{i}}{\partial t}}{\frac {\partial u^{j}}{\partial x^{k}}}-{\frac {\partial u^{j}}{\partial t}}{\frac {\partial u^{i}}{\partial x^{k}}}\right){\frac {d\gamma ^{k}}{ds}}\right]dtds\right)\\&=&\int _{\alpha _{i_{0}}}^{\alpha _{i_{0}+1}}\left.\omega _{ij}\left({\frac {\partial u^{i}}{\partial t}}{\frac {\partial u^{j}}{\partial x^{k}}}-{\frac {\partial u^{j}}{\partial t}}{\frac {\partial u^{i}}{\partial x^{k}}}\right)\right|_{t=0}{\frac {d\gamma ^{k}}{ds}}ds\\&=&\int _{\alpha _{i_{0}}}^{\alpha _{i_{0}+1}}\omega _{ij}\left(a^{i}\delta _{jk}-a^{j}\delta _{ik}\right){\frac {d\gamma ^{k}}{ds}}ds\\&=&\int _{\alpha _{i_{0}}}^{\alpha _{i_{0}+1}}(a^{i}\omega _{ik}-a^{j}\omega _{kj}){\frac {d\gamma ^{k}}{ds}}ds\\&=&\int _{\alpha _{i_{0}}}^{\alpha _{i_{0}+1}}a^{i}(\omega _{ik}-\omega _{ki}){\frac {d\gamma ^{k}}{ds}}ds\end{aligned}}}
On the other hand
{\displaystyle {\begin{aligned}X\lrcorner (\omega \phi _{i_{0}j_{0}})&=&a^{i}{\frac {\partial }{\partial x^{i}}}\lrcorner (\omega _{kl}dx^{k}\wedge dx^{l})\\&=&a^{i}\omega _{il}dx^{l}-a^{i}\omega _{ki}dx^{k}\\&=&a^{i}(\omega _{ik}-\omega _{ki})dx^{k}\\&=&a^{i}(\omega _{ik}-\omega _{ki}){\frac {d\gamma ^{k}}{ds}}ds.\end{aligned}}}
Comparing the two expressions we obtain the desired result.

Exercise 8.7

Let ${\displaystyle SL(n,\mathbb {C} )}$ act on ${\displaystyle M_{n}(\mathbb {C} )}$ by conjugation. Show that the invariant polynomials are the symmetric polynomials in the eigenvalues. What are the semistable points of this action? What the polystable ones?

Proof

We know from linear algebra that a complete invariant for this action is the Jordan form, i.e. any ${\displaystyle G=SL(n,\mathbb {C} )}$-orbit contains a unique matrix in Jordan form.

Combine this with the observation that diagonalizable matrices are Zariski-dense. A ${\displaystyle G}$-invariant polynomial is therefore determined by its value on diagonal matrices. Their entries coincide with the eigenvalues in this case. Observe that permutation matrices lie in ${\displaystyle G}$ and they just change the order of diagonal entries; we conclude that our ${\displaystyle G}$-invariant polynomial must be a symmetric polynomial in the eigenvalues.

Notice that the coefficients of the characteristic polynomial are regular functions on the space of matrices (i.e. polynomials in their entries) and that they are precisely the elementary symmetric functions in the eigenvalues. It is known that they constitute a free basis of the algebra of symmetric polynomials. So ${\displaystyle \mathbb {C} [M_{n\times n}]^{G}=\mathbb {C} [\sigma _{1},\ldots ,\sigma _{n}]}$ and the GIT quotient is ${\displaystyle \mathbb {A} _{\mathbb {C} }^{n}}$.

From this we know that unstable points are nilpotents matrices (i.e. the ones whose eigenvalues are all zero). It is maybe interesting to remark the fact that the GIT quotient is not at all an orbit space in this case, since matrices with same eigenvalues but different Jordan forms get identified in the quotient, even though their orbits are distinct.

We want now to understand the relation between orbits which get identified in the quotient. The matrix ${\displaystyle \left({\begin{smallmatrix}\epsilon &0\\0&\epsilon ^{-1}\end{smallmatrix}}\right)}$ conjugates ${\displaystyle \left({\begin{smallmatrix}\lambda &1\\0&\lambda \end{smallmatrix}}\right)}$ into ${\displaystyle \left({\begin{smallmatrix}\lambda &\epsilon ^{2}\\0&\lambda \end{smallmatrix}}\right)}$. Roughly speaking, this shows that a matrix with a shorter Jordan block lies in the closure of the orbit of a matrix with a longer Jordan block. Hence the polystable orbits (i.e. the closed ones, i.e. the smallest ones) are precisely the ones with no 1s' on the upper diagonal.

Similarly we are convinced that the largest orbits (i.e. the ones with stabilizer as small as possible, i.e. the ones corresponding to stable points) are the ones with as many 1s' as possible. To se this properly, concentrate on the stabilizer instead. If there are two different eigenvalues, then of course their eigenspaces cannot be interchanged; so we are reduced to the case where there is only one eigenvalue, and we may suppose that it is 0 since scalar matrices are central. If we look at nilpotent matrices, ${\displaystyle G}$-orbits (or, equivalently, Jordan forms) are in one-to-one correspondance with partitions of ${\displaystyle n}$, say ${\displaystyle {\underline {\lambda }}=(0\leq \lambda _{1}\leq \ldots \leq \lambda _{n}\leq n)}$. An element ${\displaystyle g\in G}$ stabilizes ${\displaystyle {\underline {\lambda }}}$ if and only if ${\displaystyle g\colon \mathbb {C} ^{n}\rightarrow \mathbb {C} ^{n}}$ is of ${\displaystyle \mathbb {C} [t]}$-modules, where ${\displaystyle t}$ acts as ${\displaystyle X_{\lambda }}$, the Jordan matrix corresponding to ${\displaystyle \lambda }$, so that ${\displaystyle \mathbb {C} ^{n}}$ splits as ${\displaystyle \bigoplus _{i=1}^{n}\mathbb {C} [t]/(t^{\lambda _{i}})}$. So indeed the largest orbits are the ones with as many 1s' as possible in their Jordan forms.

Exercise 8.8

Consider the action of ${\displaystyle B\in U(k)}$ on ${\displaystyle A\in Hom(\mathbb {C} ^{k},\mathbb {C} ^{n})}$ given by ${\displaystyle A\mapsto AB}$. Show that this has moment map ${\displaystyle A\mapsto {\frac {i}{2}}A^{\dagger }A\in u(k)}$, where u(k) consists of skew-Hermitian matrices and we are implicitly identifying ${\displaystyle u(k)}$ with its dual via the inner product ${\displaystyle X.Y=\operatorname {Tr} }$. Identify the Grassmannian ${\displaystyle Gr(k,n)}$ of complex k-planes in ${\displaystyle \mathbb {C} ^{n}}$ with the symplectic quotient ${\displaystyle \mu ^{-1}(iI)/U(k)}$.

Proof

We do not want to get bogged down in indices, so we use a trick. Notice the group acts linearly on the complex vector space ${\displaystyle {\text{Hom}}(\mathbb {C} ^{k},\mathbb {C} ^{n})}$, which is implicitly given some Hermitian structure. So we need a lemma:

Take a unitary linear representation ${\displaystyle \rho :u(k)\rightarrow End(V)}$ on a Hermitian vector space ${\displaystyle (V,J,\omega )}$ and fix an element B in ${\displaystyle u(k)}$. The Hamiltonian ${\displaystyle H:V\rightarrow \mathbb {R} }$ of the associated ${\displaystyle u(1)}$-action generated by B is given by ${\displaystyle H(x)=-1/2}$ for ${\displaystyle x\in V}$.

Proof

${\displaystyle dH=-i_{\rho (B)}(\omega )=-\omega (\rho (B)x,\_)=-}$ This is linear in the ${\displaystyle x}$ variable, so we look for a quadratic function with this differential. The answer is just what we stated. (This implicitly uses the symmetry of the matrix defined by ${\displaystyle J\rho (B)}$)

Now notice the group acts on the right on the vector space, so when we apply the lemma we need to throw in another negative sign for the Lie algebra action. Then the Hamiltonian becomes:

${\displaystyle H(A)=1/2\operatorname {Tr} ={\frac {1}{2}}i\operatorname {Tr} }$
From which we read off the moment map as stated. This is automatically equivariant so we are done.

For the second part, notice up to scaling, we are looking at the set of matrices A such that ${\displaystyle AA^{\dagger }=I}$. This amounts to isometric embeddings of ${\displaystyle \mathbb {C} ^{k}}$ into ${\displaystyle \mathbb {C} ^{n}}$. Hence up to U(k) equivalence we are just looking at embedded copies of k-dimensional subspaces, hence the symplectic quotient is naturally a Grassmannian.