Let $A: \mathbb{R}^n \to \mathbb{R}^n$ be an $\mathbb{R}$-linear map. If $1 \leq k \leq n$, we consider

$\bigwedge^k A: \bigwedge^k \mathbb{R}^n \to \bigwedge^k \mathbb{R}^n $.

What is a simple representation-theoretic argument that $\det( \bigwedge^k A ) = \det(A)^\binom{n-1}{k-1}$ ?

This is I think a formula by Sylvester, if I remember correctly. Can we say perhaps that the map $A \mapsto \det( \bigwedge^k A )$ is $GL(n)$-invariant (and maybe something else) so it must be a power of $\det(A)$?

I know it is ``elementary'', but I would like the precise proposition please.

  • 1
    $\begingroup$ Lemma 5.18 in cip.ifi.lmu.de/~grinberg/algebra/dkr1978.pdf justifies your argument (not just over $\mathbb{R}$ but over any infinite field). Note that the $\operatorname{GL}\left(n\right)$-invariance we must use here is a kind of invariance with respect to the action of $\operatorname{GL}\left(n\right)$ by matrix multiplication, not to the action by conjugation (since the latter has too many other invariants). It is not literal invariance, but rather "$f\left(BX\right) = c_B f\left(X\right)$ where $c_B$ is some scalar depending only on $B$". $\endgroup$ – darij grinberg Mar 1 '19 at 18:57
  • $\begingroup$ @darij grinberg, thank you. Yes, this is the lemma I wanted. $\endgroup$ – Malkoun Mar 1 '19 at 19:54

First of all we can clearly focus on invertible matrices (if $A$ is not invertible, $\bigwedge^k A$ isn't either for $k$ within the given bounds, so the determinant is $0$).

Then, for these notice that $f: A\mapsto \det (\bigwedge^k A)$ is a continuous group morphism $GL_n(\mathbb{R})\to \mathbb{R}^\times$, in particular since $GL_n(\mathbb{R})' = SL_n(\mathbb{R})$ and $\det$ realizes an (topological group) isomorphism $GL_n(\mathbb{R})/SL_n(\mathbb{R}) \to \mathbb{R}^\times$, it follows that $f$ factors through the determinant.

So $f = g\circ \det$ for some continuous morphism $g : \mathbb{R}^\times \to\mathbb{R}^\times$. But one can easily check that the only such continuous morphisms are of the form $x\mapsto \epsilon(x)^i |x|^l$ for some $i\in \{0,1\}$ and $l$ a real number, where $\epsilon(x)$ denotes the sign of $x$, i.e. $+/- 1$.

Therefore $f(A) = \epsilon(\det(A))^i |\det(A)|^l$ for some $i,l$ as above. Now check on some well-chosen matrices to find what $i$ and $l$ are (if $f(A)<0$ for some $A$, $\epsilon = 1$, otherwise $\epsilon = 0$, and then $l$ should follow by checking on a homotethy and getting a suitable power of the factor)

I don't know if this is "representation-theoretic"

  • $\begingroup$ Yes that works. Thank you! So the key observation is that it is a group homomorphism from $GL(n, \mathbb{C})$ onto $\mathbb{C}^*$, which is trivial on $SL(n, \mathbb{C})$, itself being the commutator subgroup of $GL(n,\mathbb{C})$. Then it follows (note that I am working over $\mathbb{C}$ for simplicity), that it must be an integer power of the determinant. Did I get the last part right, when working over $\mathbb{C}$? $\endgroup$ – Malkoun Feb 28 '19 at 15:49
  • $\begingroup$ $\mathbb{C}$ doesn't make things simpler actually, because then a continuous morphism $\mathbb{C}^\times \to \mathbb{C}^\times$ can look like $z\mapsto |z|^\alpha (\frac{z}{|z|})^ke^{i\beta \ln |z|}$ for an integer $k$ and reals $\alpha,\beta$, so you have to find som examples again and determine $\alpha, \beta, k$ - though of course by taking $A$ to be a unital homotethy you can get $k$, then by taking $A$ to be real with positive determinant you can get $\alpha, \beta$. (But you are right about the key observation) $\endgroup$ – Maxime Ramzi Feb 28 '19 at 16:04
  • $\begingroup$ You are right. But one may also take into account in the proposition homogeneity under complex scaling, which is satisfied in my example. Then we get an integer power of the determinant. $\endgroup$ – Malkoun Feb 28 '19 at 16:25
  • $\begingroup$ That's right ! It works that way $\endgroup$ – Maxime Ramzi Feb 28 '19 at 16:47

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