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

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

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


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.


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 $g_1=g_2$.

  • $\begingroup$ 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

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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