When the affine quotient is faithfully flat? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T19:52:45Z http://mathoverflow.net/feeds/question/104032 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/104032/when-the-affine-quotient-is-faithfully-flat When the affine quotient is faithfully flat? hoxide 2012-08-05T16:19:23Z 2012-08-06T19:15:35Z <p>It may be easy for the expert. </p> <p>Consider the map from $n$ by $m$ matrices (over $\mathbb{C}$ )to the $n$ by $n$ symmetric matrices $\phi\colon A\mapsto A A^T$. My question is when this map is faithfully flat? </p> <p>I known that for the faithfully part, (if already known it is flat) it is equivalent to the surjectivity of $\phi$. So the condition should at least be $m\geq n$. </p> <p>I also known that when $m\geq 2n$, $\mathbb{C}[M_{n,m}] = \mathbb{C}[AA^T] \otimes H$ where $H$ is called Harmonics. So it is true for $m\geq 2n$. </p> <p>I guess the map $\phi$ is faithfully flat exactly for $m\geq n$. </p> <p>I also ask the similar questions for map $\phi\colon M_{n,r}\times M_{m,r}\to M_{n,m}$ by $(A,B)\mapsto AB^T$ and map $\phi\colon M_{2n, m} \to Alt_m$ by $A\mapsto A^TJA$ where $Alt_m$ is the space of antisymmetric matrices and $J$ is a symplectic form. The situations are similar, I guess the conditions are exactly the conditions for $\phi$ to be surjective, while I known the claim is true for a stronger condition. </p> <p>All comments are welcome! Thank you very much!</p> http://mathoverflow.net/questions/104032/when-the-affine-quotient-is-faithfully-flat/104139#104139 Answer by Torsten Wedhorn for When the affine quotient is faithfully flat? Torsten Wedhorn 2012-08-06T19:15:35Z 2012-08-06T19:15:35Z <p>The morphism $\phi$ is not flat for $n = m = 2$. Otherwise all fibers would be of dimension 1, namely the dimension of all 2 by 2 matrices minus the dimension of all symmetric 2 by 2 matrices. But the fiber over the zero matrix has dimension 2: matrices with coefficients $a$, $b = ia$, $c$, $d = ic$ for $i$ a square root of -1 form a 2-dimensional irreducible component in the zero fiber (in fact it is a pleasant exercise to check that the zero fiber has 6 irreducible components, two of them of dimension 2 and four of dimension 1).</p> <p>If you restrict $\phi$ to the invertible $n$ by $n$ matrices, then $\phi$ is flat (even smooth) and its image consists of all invertible symmetric matrices. In fact, as source and target of $\phi$ are smooth over the base field it suffices to show that the induced map on tangent spaces is surjective. This can be checked directly (at the invertible $n$ by $n$ matrix $A$ the tangent map is given by $X \mapsto AX^t + XA^t$ and it is not difficult to check that this map from all $n$ by $n$ matrices to the space of symmetric $n$ by $n$ matrices is surjective) or you can argue as follows: $GL_n$ acts transitively on the set of invertible symmetric matrices $S$ by $(g,S) \mapsto gSg^t$ (over any algebraically closed field of characteristic $\ne 2$) and on the set of invertible $n$ by $n$ matrices $A$ by left multiplication, and $\phi$ is equivariant for this action. Thus it suffices to check the surjectivity of the tangent map in one point. But for $A = I_n$ this is trivial.</p>