Generators for the algebra of GL(n)-equivariant maps from M_n + M_n to M_n - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T12:59:17Z http://mathoverflow.net/feeds/question/39194 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/39194/generators-for-the-algebra-of-gln-equivariant-maps-from-m-n-m-n-to-m-n Generators for the algebra of GL(n)-equivariant maps from M_n + M_n to M_n Bart Van Steirteghem 2010-09-18T03:09:12Z 2010-09-19T10:34:49Z <p>Let $M_n$ be the set of $n$-by-$n$ matrices with complex entries, viewed as a variety over $k=\mathbb{C}$. Equip $M_n$ with the conjugation action of $\mathrm{GL}(n)=\mathrm{GL}(n,\mathbb{C})$. Consider $A:=\mathrm{Mor}^{\mathrm{GL}(n)}(M_n \oplus M_n, M_n)$, the set of $\mathrm{GL}(n)$-equivariant maps (of algebraic varieties) $M_n \oplus M_n \to M_n$, where $GL(n)$ acts diagonally on $M_n \oplus M_n$. Then the (standard) algebra structure of $M_n$ (with multiplication given by matrix multiplication) induces an algebra structure on $A$. </p> <p>The following maps belong to $A$:</p> <ol> <li>$M_n \oplus M_n \to M_n \colon (A,B) \mapsto A$;</li> <li>$M_n \oplus M_n \to M_n \colon (A,B) \mapsto B$;</li> <li>$M_n \oplus M_n \to M_n \colon (A,B) \mapsto f(A,B)I_n$, where $I_n$ is the identity matrix and $f$ is an element of the ring of invariants $k[M_n \oplus M_n]^{\mathrm{GL}(n)}$. </li> </ol> <p>Do they generate $A$ as an algebra?</p> http://mathoverflow.net/questions/39194/generators-for-the-algebra-of-gln-equivariant-maps-from-m-n-m-n-to-m-n/39203#39203 Answer by Andreas Thom for Generators for the algebra of GL(n)-equivariant maps from M_n + M_n to M_n Andreas Thom 2010-09-18T07:43:44Z 2010-09-18T10:43:17Z <p>If you consider pairs of unitaries instead, and the group, that you get out of a similar construction, the answer is negative. That is an analogous question but in a slightly different context. The new question is only interesting if one studies it for all dimensions at once.</p> <p>To be more precise, in <a href="http://arxiv.org/abs/1003.4093" rel="nofollow">http://arxiv.org/abs/1003.4093</a>, it was shown that there are exotic families of continuous maps $$\phi_n \colon U(n) \times U(n) \to U(n),$$ such that:</p> <p>1) $\phi_{n+m}(U \oplus V,U' \oplus V') = \phi_n(U,V) \oplus \phi_n(U',V')$,</p> <p>2) $\phi_{nm}(U \otimes V,U' \otimes V') = \phi_n(U,V) \otimes \phi_n(U',V')$, and</p> <p>3) $\phi_n(AUA^{-1},AVA^{-1}) = A\phi_n(U,V)A^{-1}$.</p> <p>Here, exotic means that $\phi_n(U,V)$ is not given by evaluating the pair of unitaries at a word $w \in {\mathbb F}_2$, where ${\mathbb F}_2$ denotes the free group on two generators.</p> http://mathoverflow.net/questions/39194/generators-for-the-algebra-of-gln-equivariant-maps-from-m-n-m-n-to-m-n/39291#39291 Answer by Victor Protsak for Generators for the algebra of GL(n)-equivariant maps from M_n + M_n to M_n Victor Protsak 2010-09-19T10:34:49Z 2010-09-19T10:34:49Z <p>The answer is affirmative not only in the case of 2 matrices, but also in the case of any number of matrices; in fact, an analogous statement is true for quiver representations (in characteristic 0).</p> <p>The original question can be restated as follows.</p> <blockquote> <p>Let $P$ be the space of polynomial functions of 2 $n\times n$ matrices, with the adjoint action of $GL_n$ and the ring of invariants $I.$ Consider the space $\text{Hom}_{GL_n}(M_n,P)$ as an $I$-module. Is it true that it is generated by the products of matrices?</p> </blockquote> <p>For the case of any number of generic matrices $A_1,\ldots,A_k,$ Procesi proved that over a field $k$ of characteristic 0, $I$ is spanned by the traces of the products of matrices. Formally, consider words in the free monoid with $k$ generators, substitute the generic matrices, and take a trace. </p> <p>Procesi, C. <em>The invariant theory of n×n matrices</em>. Advances in Math. 19 (1976), no. 3, 306&ndash;381</p> <p>The statement follows by adjoining an extra generic matrix $A_0$ and converting an $M_n$-space into a $GL_n$-invariant forming a product with $A_0$ and taking the trace, then undoing the trace of the term in the trace polynomial from Procesi's theorem containing $A_0.$ </p> <hr> <p>Here is a vast generalization due to Le Bruyn and Procesi. Given a finite quiver $Q$ and a dimension vector $\alpha,$ consider the corresponding representation space $R(Q,\alpha)$ with the action of the algebraic group $GL(\alpha)$ and the space $P$ of polynomial functions on $R.$ (If the quiver consists of a single vertex with $k$ loops and $\alpha=n$ then the representation space is given by $k$ generic $n\times n$ matrices with the simultaneous conjugation action by $GL_n.$) Then, over a field of characteristic 0, the algebra $I$ of polynomial invariants is spanned by the traces of matrix products over oriented cycles in $Q$ and for any pair of vertices $(i,j)$ of $Q,$ the space <code>$\text{Hom}_{GL(\alpha)}(\text{Hom}_k(V_i,V_j),P)$</code> is generated as an $I$-module by the products over oriented paths connecting $i$ with $j.$</p> <p>Lieven Le Bruyn, Claudio Procesi, <em>Semisimple representations of quivers</em>. Trans. Amer. Math. Soc. 317 (1990), no. 2, 585&ndash;598</p>