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Given $n \in \mathbb N$, I am asked to show that there is a multilinear symmetric $\operatorname{GL}_n$-invariant form $\phi : (M_{n \times n})^l \to \mathbb R$ (for some $l \geq 0$) such that $\phi(A,A,...,A) = \det A$ for all $A \in \mathbb M_{n \times n}(R)$.

Using the idea of polarizing an algebraic form, I decided to do the following: by definition, we have $$ \det(A) = \sum_{\sigma \in S_n} \operatorname{sgn}(\sigma) \prod_{i=1}^n A_{i \sigma(i)} $$

and therefore we could define $\phi$ with $l=n$, but taking $n$ independent copies of $A$ and adding another permutation into the mix, just like the polarization formula. This gives: $$ \phi(A^{1},A^2,...,A^n) = \frac 1{n!} \sum_{\pi \in S_n} \sum_{\sigma \in S_n} \operatorname{sgn}(\sigma)\prod_{i=1}^n A^{\pi(i)}_{i\sigma(i)} $$

This formula is multilinear, symmetric and $\phi(A,A,..,A) = \det A$. However, I am not quite getting how to prove invariance : $\phi(g^{-1}A^1g, g^{-1}A^2g,...,g^{-1}A^ng) = \phi(A^1,A^2,...,A^n)$ for all $A^i$, $i = 1 \to n$.

While this may not work out, I am inclined to think it does, since I used the definition of polarization to obtain all the conditions, and got the one I need additionally. However, proceeding by simple expansion does not work (creating a bunch of $g^{-1}$ and $g$ indexed terms, and therefore discombobulation), and therefore I need some help on why this is the case.

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  • $\begingroup$ Is invariance really required here? $\endgroup$ – Berci Mar 17 '18 at 9:53
  • $\begingroup$ Unfortunately, it is. You see, when the determinant itself satisfies invariance, I would like to think that such a property can be satisfied by its polarization as well. I was told this is true, so it must be. $\endgroup$ – Teresa Lisbon Mar 17 '18 at 10:03
  • $\begingroup$ What about taking $\psi(A_1,\dots, A_n) :=\frac1n\big(\phi(A_1,A_2,\dots) +\phi(A_2,A_3,\dots, A_1)+\dots+\phi(A_n, A_1,\dots) \big) $? $\endgroup$ – Berci Mar 17 '18 at 10:28
  • $\begingroup$ I am not sure that will be symmetric, but am willing to check. In particular, if one switches $A^2$ and $A^3$ only, say, then what would we do? $\endgroup$ – Teresa Lisbon Mar 17 '18 at 10:34
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    $\begingroup$ @астонвіллаолофмэллбэрг: You are on the right track; your $\phi$ is invariant. The easiest way I see to prove this is to show that $\phi\left(A_1, A_2, \ldots, A_n\right)$ is the coefficient of $t_1 t_2 \cdots t_n$ in the polynomial $\left(1/n!\right) \det\left(t_1 A_1 + t_2 A_2 + \cdots + t_n A_n\right)$. (This is, of course, the old way to construct polarizations of arbitrary polynomial functions.) $\endgroup$ – darij grinberg Mar 21 '18 at 22:06

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