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

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 $\Lambda^k V$, in the following way

$$ \langle v_1 \wedge \dots \wedge v_k , w_1 \wedge \dots \wedge w_k \rangle:=\det (\langle v_i ,w_j \rangle). \tag{1}$$

Question:

What are necessary and sufficient conditions on an inner product on $\Lambda^k V$ to to be induced from a product on $V$ as in $(1)$?

(I restricted $k \neq 1,d$ because then of course, any metric is induced by a metric on the base).

I think there should be some "compatibility" or "symmetry" conditions, but I am not sure how to formulate them.


Edit:

If there exist an inducing product at the base, this product is unique. Perhaps we can construct an "inverse map" which is defined on all products on $\Lambda^k V$, and see when the result is an honest inner product on $V$.

Proof of uniqueness:

Suppose $g_1,g_2$ are inner products on $V$, which induce the same product on $\Lambda^k V$.

Let $A:(V,g_1) \to (V,g_2)$ be an isometry. Then the induced exterior map $$\bigwedge ^k A:(\Lambda^k V,\Lambda^k g_1) \to (\Lambda^k V,\Lambda^k g_2)$$

is an isometry, where $\Lambda^k g_i$ is the metric on $ \Lambda^k V$ induced by $g_i$ via $(1)$. Since by assumption $\Lambda^k g_1=\Lambda^k g_2$, we get that

$$\bigwedge ^k A:(\Lambda^k V,\Lambda^k g_1) \to (\Lambda^k V,\Lambda^k g_1)$$

is an isometry, i.e.

$$ \id_{\Lambda_k(V)}=(\bigwedge ^k A)^T \circ \bigwedge ^k A= \bigwedge ^k A^T \circ \bigwedge ^k A=\bigwedge ^k A^T A,$$

where the transpose in $A^T$ is taken with respect to the metric $g_1$. Denote $S=A^TA$. Then $S \in \Hom(V,V)$ is symmetric (w.r.t. $g_1$) and positive-definite, and satisfies $ \id_{\Lambda_k(V)} =\bigwedge^k S $.

This implies $S=\id_V$. (For a short proof, see this answer).

So, we have obtained $A^TA=\id_V$; Remembering the transpose was taken w.r.t $g_1$, this shows $A$ is an automorphism of $(V,g_1)$. Recalling that we assumed $A:(V,g_1) \to (V,g_2)$ was an isometry, this shows $g_1=g_2$.

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  • $\begingroup$ For $d=2$ this is related to the classification/decomposition of algebraic curvature tensors, so that might be a place to start reading for some ideas? (You're looking for elements of $S^2 \Lambda^2 V$ arising as the curvature tensor of some constant-curvature space.) The Bianchi identity is the obvious requirement, but the rest (traceless Ricci and vanishing Weyl) becomes trickier to identify when you don't already know the metric. $\endgroup$ – Anthony Carapetis Dec 30 '17 at 2:23
  • $\begingroup$ If $g$ is a metric on $V$ then I write the corresponding induced metric on $\Lambda^2 V$ as $g \wedge g$. (Here $\wedge$ is the Kulkarni-Nomizu product up to a constant.) In the case where $V=T_pM$ is a tangent space of a Riemannian manifold and $g$ the metric at $p$, the $(0,4)$ curvature tensor at $p$ can be interpreted as a bilinear form $R$ on $\Lambda^2 V$, and we have $R = g \wedge g$ if and only if the sectional curvatures of the metric are all equal to $1$ at $p$. $\endgroup$ – Anthony Carapetis Dec 30 '17 at 7:17
  • $\begingroup$ Thus we can rephrase your question: given a (positive-definite) $R \in S^2 \Lambda^2 V$, when is there a metric $g \in S^2 V$ such that $R,g$ together look like the geometry of a sphere? Not sure if this will help, just a thought. $\endgroup$ – Anthony Carapetis Dec 30 '17 at 7:22
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This is a partial answer, for the case $k=d-1$:

In this case, $\Lambda^{d-1} V$ and $V$ have the same dimension, so it is reasonable to expect any metric on $\Lambda^{d-1} V$ comes from a metric on $V$. We shall prove this is indeed the case.

First, we reformulate the problem:

A choice of a metric 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 induced 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 $h=\Lambda^kg$?

In the case of $k=d-1$, the answer is that all such $h$ come from $g$'s at the base.

Indeed, it follows from this question that every orientation-preserving invertible linear map $\bigwedge^{d-1} V \to \bigwedge^{d-1} V^*$ equals $\Lambda^{d-1}A$ for some invertible $A:V \to V^*$.

A metric (regarded as a map $V \to V^*$) is always orientation-preserving (it maps an orthonormal basis $e_i$ to its dual basis), so our $h$ equals $\Lambda^{d-1}g$ for some $g$. We now need to prove $g$ is symmetric and positive.

The symmetry is not hard to prove. Also, it is quite easy to see $g$ must be definite (negative or positive). If $d-1$ is odd, it must be positive-definite. When $d-1$ is even $\Lambda^{d-1}g=\Lambda^{d-1}(-g)$, so we can always choose the "positive root".

This finishes the proof.

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