Polynomials and Representations XXXIX

Some Invariant Theory

We continue the previous discussion. Recall that for \mu = \overline\lambda we have a GL_n\mathbb{C}-equivariant map

\displaystyle \bigotimes_j \text{Alt}^{\mu_j} \mathbb{C}^n \to \bigotimes_i \text{Sym}^{\lambda_i} \mathbb{C}^n \subset\mathbb{C}[z_{i,j}]^{(\lambda_i)}, \quad e_T^\circ \mapsto D_T

which induces an isomorphism between the unique copies of V(\lambda) in both spaces. The kernel Q of this map is spanned by e_T^\circ - \sum_S e_S^\circ for various fillings T with shape \lambda and entries in [n].

E.g. suppose \lambda = (4, 3, 1) and n=5; then \mu = (3, 2, 2,1) and the map induces:

\displaystyle \text{Alt}^3 \mathbb{C}^5 \otimes \text{Alt}^2 \mathbb{C}^5 \otimes \text{Alt}^2 \mathbb{C}^5\otimes \mathbb{C}^5 \longrightarrow \mathbb{C}[z_{1,j}]^{(4)} \otimes \mathbb{C}[z_{2,j}]^{(3)}\otimes \mathbb{C}[z_{3,j}]^{(1)}, \ 1\le j \le 5,

with kernel Q. For the following filling T, we have the correspondence:

example_determinant_and_alt

Lemma. If \mu_j = \mu_{j+1} = \ldots = \mu_{j+m-1}, then the above map factors through

\displaystyle \text{Alt}^{\mu_j}\mathbb{C}^n \otimes \ldots \otimes \text{Alt}^{\mu_{j+m-1}}\mathbb{C}^n\longrightarrow \text{Sym}^m \left(\text{Alt}^{\mu_j} \mathbb{C}^n\right).

E.g. in our example above, the map factors through \text{Alt}^3 \mathbb{C}^5 \otimes \text{Sym}^2 \left(\text{Alt}^2 \mathbb{C}^5 \right)\otimes \mathbb{C}^5.

Proof

Indeed if \mu_j = \mu_{j+1}, then swapping columns j and j+1 of T gives us the same D_T. ♦

blue-lin

Now suppose 0\le a\le n and m\ge 0; pick \lambda = (m, \ldots, m) with length a. Now \mu comprises of m copies of a so by the above lemma, we have a map:

\displaystyle \text{Sym}^m \left(\text{Alt}^a \mathbb{C}^n\right) \longrightarrow \mathbb{C}[z_{1,j}]^{(m)} \otimes \ldots \otimes \mathbb{C}[z_{a,j}]^{(m)} \cong \mathbb{C}[z_{i,j}]^{(m,\ldots,m)}

where 1 \le i \le a and 1 \le j \le n. Taking the direct sum over all m we have:

\displaystyle \text{Sym}^* \left(\text{Alt}^a \mathbb{C}^n\right) := \bigoplus_{m\ge 0} \text{Sym}^m \left(\text{Alt}^a \mathbb{C}^n\right)\longrightarrow \bigoplus_{m\ge 0}\mathbb{C}[z_{i,j}]^{(m,\ldots,m)} \subset \mathbb{C}[z_{i,j}]

which is a homomorphism of GL_n\mathbb{C}-representations. Furthermore, \text{Sym}^* V for any vector space V has an algebra structure via \text{Sym}^m V \times \text{Sym}^n V \to \text{Sym}^{m+n}V. The above map clearly preserves multiplication since multiplying e_T^\circ e_{T'}^\circ and D_T D_{T'} both correspond to concatenation of T and T’. So it is also a homomorphism of \mathbb{C}-algebras.

Question. A basis of \text{Alt}^a\mathbb{C}^n is given by

\displaystyle e_{i_1, \ldots, i_a} := e_{i_1} \wedge \ldots \wedge e_{i_a}

for 1 \le i_1 < \ldots < i_a \le n. Hence \text{Sym}^*\left(\text{Alt}^a \mathbb{C}^n\right) \cong \mathbb{C}[e_{i_1,\ldots,i_a}], the ring of polynomials in n\choose a variables. What is the kernel P of the induced map:

\mathbb{C}[e_{i_1, \ldots, i_a}] \longrightarrow \mathbb{C}[z_{1,1}, \ldots, z_{a,n}]?

Answer

We have seen that for any 1 \le i_1 < \ldots < i_a \le n and 1 \le j_1 < \ldots < j_a \le n we have:

\displaystyle e_{i_1, \ldots, i_a} e_{j_1, \ldots, j_a} = \sum e_{i_1', \ldots, i_a'} e_{j_1', \ldots, j_k', j_{k+1},\ldots, j_a}

where we swap j_1, \ldots, j_k with various sets of k indices in i_1, \ldots, i_a while preserving the order to give (i_1', \ldots, i_a') and (j_1', \ldots, j_k'). Hence, P contains the ideal generated by all such quadratic relations.

On the other hand, any relation e_T = \sum_S e_S is a multiple of such a quadratic equation with a polynomial. This is clear by taking the two columns used in swapping; the remaining columns simply multiply the quadratic relation with a polynomial. Hence P is the ideal generated by these quadratic equations. ♦

Since the quotient of \mathbb{C}[e_{i_1, \ldots, i_a}] by P is a subring of \mathbb{C}[z_{i,j}], we have:

Corollary. The ideal generated by the above quadratic equations is prime.

blue-lin

Fundamental Theorems of Invariant Theory

Recall that g = (g_{i,j}) \in GL_n\mathbb{C} takes z_{i,j} \mapsto \sum_k z_{i,k} g_{k,j}, which is a left action; here 1\le i\le a, 1\le j\le n. We also let GL_a\mathbb{C} act on the right via:

\displaystyle g=(g_{i,j}) \in GL_a\mathbb{C} :z_{i,j} \mapsto \sum_k g_{i,k}z_{k,j}

so that \mathbb{C}[z_{i,j}] becomes a (GL_n\mathbb{C}, GL_a\mathbb{C})-bimodule. A basic problem in invariant theory is to describe the ring \mathbb{C}[z_{i,j}]^{SL_a\mathbb{C}} comprising of all f such that g\cdot f = f for all g\in SL_a\mathbb{C}.

Theorem. The ring \mathbb{C}[z_{i,j}]^{SL_a\mathbb{C}} is the image of:

\displaystyle \mathbb{C}[D_{i_1, \ldots, i_a}] \cong \mathbb{C}[e_{i_1, \ldots, i_a}]/P \hookrightarrow \mathbb{C}[z_{i,j}]

where (i_1, \ldots, i_a) runs over all 1 \le i_1 < \ldots < i_a \le n.

In other words, we have:

  • First Fundamental Theorem : the ring of SL_a\mathbb{C}-invariants in \mathbb{C}[z_{i,j}] is generated by \{D_{i_1, \ldots, i_a}: 1 \le i_1 < \ldots < i_a \le n\}.
  • Second Fundamental Theorem : the relations satisfied by these polynomials are generated by the above quadratic relations.

blue-lin

Proof of Fundamental Theorems

Note that g takes D_{i_1, \ldots, i_a} to:

\displaystyle \det{\small \begin{pmatrix} z_{1, i_1} & z_{1, i_2} & \ldots & z_{1, i_a} \\ z_{2, i_1} & z_{2, i_2} & \ldots & z_{2, i_a} \\ \vdots & \vdots & \ddots & \vdots \\ z_{a, i_1} & z_{a, i_2} & \ldots & z_{a, i_a}\end{pmatrix}} \mapsto \det {\small \begin{pmatrix}\sum_{j_1} g_{1,j_1} z_{j_1, i_1} & \ldots & \sum_{j_a} g_{1,j_a} z_{j_a, i_a} \\ \sum_{j_1} g_{2,j_1} z_{j_1, i_1} & \ldots & \sum_{j_a} g_{2,j_a} z_{j_a, i_a} \\ \vdots & \ddots & \vdots \\ \sum_{j_1} g_{a,j_1} z_{j_1, i_1} & \ldots & \sum_{j_a} g_{a,j_a} z_{j_a, i_a} \end{pmatrix}} = \det(g) D_{i_1, \ldots, i_a}

which is D_{i_1, \ldots, i_a} if g \in SL_a\mathbb{C}. Hence we have \mathbb{C}[D_{i_1, \ldots, i_a}] \subseteq \mathbb{C}[z_{i,j}]^{SL_a\mathbb{C}}. To prove equality, we show that their dimensions in degree d agree. By the previous article, the degree-d component of \mathbb{C}[e_{i_1, \ldots, i_a}]/P has a basis indexed by SSYT of type \lambda = (\frac d a, \ldots, \frac d a) and entries in [n]; if d is not a multiple of a, the component is 0.

Next we check the degree-d component of \mathbb{C}[z_{i,j}]^{SL_a\mathbb{C} }. As GL_a\mathbb{C}-representations, we have

\displaystyle \mathbb{C}[z_{i,j}] \cong \text{Sym}^*\left( (\mathbb{C}^a)^{\oplus n}\right)

where GL_a\mathbb{C} acts on \mathbb{C}^a canonically. Taking the degree-d component, once again this component is 0 if d is not a multiple of a. If a|d, it is the direct sum of \text{Sym}^{d_1} \mathbb{C}^a \otimes \ldots \otimes \text{Sym}^{d_n}\mathbb{C}^a over all d_1 + \ldots + d_n = \frac d a. The \psi of this submodule is h_{\lambda'} = \sum_{\mu'\vdash \frac d a} K_{\mu'\lambda'} s_{\mu'} where \lambda' is the sequence (d_1, \ldots, d_n). Hence

\displaystyle \text{Sym}^{d_1} \mathbb{C}^a \otimes \ldots \otimes \text{Sym}^{d_n}\mathbb{C}^a \cong \bigoplus_{\mu'\vdash \frac d a} V(\mu')^{\oplus K_{\mu'\lambda'}}.

Fix \mu' and sum over all (d_1, \ldots, d_n); we see that the number of copies of V(\mu') is the number of SSYT with shape \mu' and entries in [n]. The key observation is that each V(\mu') is an SL_a\mathbb{C}-irrep.

  • Indeed, \mathbb{C}^* \subset GL_a\mathbb{C} acts as a constant scalar on the whole of V(\mu') since \psi_{V(\mu')} = s_{\mu'} is homogeneous. Hence any SL_a\mathbb{C}-invariant subspace of V(\mu') is also GL_a\mathbb{C}-invariant.

Hence V(\mu')^{SL_a\mathbb{C}} is either the whole space or 0. From the proposition here, it is the whole space if and only if \mu' = (\frac d a, \ldots, \frac d a) with a terms (which corresponds to \det^{d/a}). Hence, the required dimension is the number of SSYT with shape (\frac d a, \ldots) and entries in [n]. ♦

blue-lin

This entry was posted in Uncategorized and tagged , , , , , . Bookmark the permalink.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s