5
$\begingroup$

$\newcommand{\id}{\text{id}}$ $\newcommand{\Hom}{\text{Hom}}$

This is a cross-post. Let $V$ be a $d$-dimensional real vector space, and let $2 \le k \le d-1$. Every inner product on $V$ induces an inner product on $\Lambda^k V$:

$$ \langle v_1 \wedge \dots \wedge v_k , w_1 \wedge \dots \wedge w_k \rangle:=\det (\langle v_i ,w_j \rangle). $$

What are necessary and sufficient conditions on a product on $\Lambda^k V$ to to be induced from a product on $V$?

For $k=d-1$ the answer is that every product on $\Lambda^{d-1} V$ is induced from a product on $V$.

Edit 1:

If there exist an inducing product at the base, this product is unique (details are provided under the "edit" here). Perhaps we can construct an "inverse map" which is defined on the space of products on $\Lambda^k V$, and see when the result is an honest inner product on $V$ (and not just a bilinear form).

Edit 2:

Here is an equivalent formulation of the question:

A choice of a product on $V$ is equivalent to a choise of a linear isomorphism $ g:V \to V^*$ that satisfies

$$ g(v)(w)=g(w)(v) \, \, \text{and}\, \,g(v)(v) \ge 0 \, \, \text{with equality only when } \, v=0. \tag{1}$$

The equivalence is via $g(v)(w):= \langle v,w \rangle$. Using this perspective, the metric on $\Lambda^{k} V$ induced by $g$ is $\Lambda^kg:\Lambda^{k} V \to \Lambda^{k} (V^*) \cong (\Lambda^{k} V)^*$.

So, the question becomes the following:

For which maps $h:\Lambda^{k} V \to (\Lambda^{k} V)^*$ which are symmetric and positive in the sense of $(1)$, there exist a symmetric and positive $g$ such that $h=\Lambda^kg$?

(The symmetry and positivity requirements on $g$ are in fact redundant-if there exist a "root" $g$ such that $h=\Lambda^kg$, then $g$ is symmetric and can be taken to be positive definite).

Peter Michor suggests using Plucker relations as a necessary condition. These relations give an equivalent conditions for an element $h\in \Lambda^k (V^* \otimes V^*)$ to be decomposable, i.e. $h=g_1\wedge g_2\wedge\dots\wedge g_k$, where $g_i \in V^* \otimes V^*$.

However, in the formulation above, $ g:V \to V^*$, and $\Lambda^kg:\Lambda^{k} V \to \Lambda^{k} (V^*) \cong (\Lambda^{k} V)^*$ is the induced map on exterior powers, that is an element of $\Lambda^{k} V^* \otimes \Lambda^{k}V^*$. Thus, in order to use the Plucker relations, we need a way to consider it as an element in $\Lambda^{k}(V^* \otimes V^*)$. Moreover, we need that this identification will map "power-elements" to "decomposable elements".

I am not sure such a map exists. Here is a precise question on this.


Yet another equivalent formulation of the question...:

Given $(\binom {d}k)^2$ numbers, indexed by ordered pairs $\big((i_1,\dots,i_k),(j_1,\dots,j_k)\big)$ where $1 \le i_1 <i_2 < \dots<i_k\le d$, under what conditions do they form the $k$ minors of some $d \times d$ matrix? i.e.

$$ b_{(i_1,\dots,i_k),(j_1,\dots,j_k)}=k-\text{minor of a } d \times d \text{ matrix, corresponding to rows } (i_1,\dots,i_k), \text{and to colums} (j_1,\dots,j_k) $$

The equivalence is obtained by choosing a fixed basis $e_1,\dots,e_d$ for $V$, and setting $$b_{(i_1,\dots,i_k),(j_1,\dots,j_k)}=\langle e_{i_1} \wedge \dots \wedge e_{i_k} , e_{j_1} \wedge \dots \wedge e_{j_k} \rangle.$$

Technically, we should also take care of the symmetry and positivity; however, it turns out that if the "upper matrix" $b$ is "symmetric*" and positive, then the underlying matrix $A$ (if exists) is symmetric and definite, and can always be chosen to be positive.

*The symmetry of the "matrix $b$" is $b_{(i_1,\dots,i_k),(j_1,\dots,j_k)}=b_{(j_1,\dots,j_k),(i_1,\dots,i_k)}$. The positivity corresponds to $b_{(i_1,\dots,i_k),(i_1,\dots,i_k)} >0$.

$\endgroup$
2
$\begingroup$

A necessary condition are the Pluecker relations. Namely, let $W= V^*\otimes V^*\ni g$, then $h=\Lambda^k g$ is decomposable. In detail, $h\in \Lambda^k W$ is decomposable, i.e., $h=g_1\wedge g_2\wedge\dots\wedge g_k$, if and only $i_{\Phi}h\wedge h = 0$ for all $\Phi\in\Lambda^{k-1}W^* = \Lambda^{k-1}(V\otimes V)$. It then remains to ensure that all $g_1$ are the same and are positive definite. See here for various equivalent versions of the Pluecker relations.

$\endgroup$
  • $\begingroup$ Thanks, though I think the link you provided doesn't work (at least for me). $\endgroup$ – Asaf Shachar Jan 11 '18 at 7:39
  • $\begingroup$ Sorry, one more question: In my formulation $g:V \to V^*$, and $\Lambda^kg:\Lambda^{k} V \to \Lambda^{k} (V^*) \cong (\Lambda^{k} V)^*$ is the induced map on exterior powers, that is an element of $\Lambda^{k} V^* \otimes \Lambda^{k} V^*$. You seem to consider it as an element in $\Lambda^{k}(V^* \otimes V^*)$. Can you please say how exactly do you identify $\Lambda^{k} V^* \otimes \Lambda^{k} V^*$ as a subset of $\Lambda^{k}(V^* \otimes V^*)$? Thanks. $\endgroup$ – Asaf Shachar Jan 11 '18 at 8:08
  • $\begingroup$ Link corrected. I did not work out the injection. The most natural one is: $(v_1^*\wedge\dots\wedge v_k^*)\otimes (w_1^*\wedge \dots\wedge w_k^*) \mapsto \pm (v_1^*\otimes w_1^*)\wedge \dots \wedge (v_k^*\otimes w_k^*)$. One should look at the corresponding Young tableaus and identify the irreducible components as $GL(V)$-representations. $\endgroup$ – Peter Michor Jan 11 '18 at 9:54
  • $\begingroup$ Thanks. This was also my natural guess, but that doesn't seem to be well-defined (consider what happens when you switch $v_1^*,v_2^*$). In fact, there is a possibility that there is no injection at all: see here. $\endgroup$ – Asaf Shachar Jan 11 '18 at 10:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.