# Inner Products on Exterior Powers

Let $H$ is a real, $n$-dimensional vector space.

Define $\varphi \colon \operatorname{GL}(H) \rightarrow \operatorname{GL}(\wedge^{k}H)$ by $A \mapsto \wedge^{k}A$ and $\psi_{\langle \cdot, \cdot \rangle} \colon \operatorname{GL}(\wedge^{k}H) \rightarrow T$ given by $J \mapsto\langle J(\cdot),J(\cdot)⟩$.

Here, $\langle \cdot, \cdot \rangle$ is some inner product on $H$, $$T = \{g \colon \wedge^{k}H\times \wedge^{k}H \, | \, g \text{ is a inner product on } \wedge^{k}H\}$$ and

$$\langle u_{1}\wedge...\wedge u_{k},v_{1}\wedge...\wedge v_{k} \rangle \colon= \det(\langle u_{i},v_{j}\rangle)_{k\times k}.$$

Define $S:=\{h:H\times H|$ $h$ is a inner product in $H\}$ and set $R_{h}:=\psi_{h} \circ \varphi(Gl(H))$.

Does it hold that $T = \bigcup_{h \in S} R_{h}$?

• No, the same technique used in the answer of the linked question can show it. Construct a map from an $\frac{n(n+1)}{2} + n^2$ dimensional manifold to $T$ whose image is your union and show that it cannot be onto by dimensional considerations. – levap Dec 26 '16 at 21:09
• Yeah. Your function (which is not linear, BTW) is not injective but this makes the problem even worse. A metric on $\Lambda^k(H)$ depends on $\Theta(n^{2k})$ (at least as long as $k \leq \frac{n}{2}$) parameters while you hope to describe it only using $\Theta(n^2)$ parameters so this cannot work in general. I encourage you to write a complete and rigourous answer to your question and accept it. – levap Dec 26 '16 at 21:28
• Well, maybe my asymptotic is wrong in general but it is definitely true if $k = 2$ (then the dimension of $\Lambda^k(H)$ is $\frac{n(n-1)}{2}$ and the dimension of the space of metrics on $\Lambda^k(H)$ is $\frac{\frac{n(n-1)}{2} \left( \frac{n(n-1)}{2} + 1\right)}{2} = \Theta(n^4)$) which is bad enough. – levap Dec 26 '16 at 21:36
• I didn't say that it always works, I just said that it cannot work for general $n,k$. I'm also quite sure it won't work even when $n = 4$ and $k = 2$ but you'll have to take into consideration the redundancy of your description. The map $\gamma$ is very not one-to-one and if you'll eliminate it and describe it more concisely using less parameters, you'll get a better estimate. If you want, I can try and show why it doesn't work even when $k = 2$ and $n = 4$. – levap Dec 26 '16 at 22:26
• I've added an answer that handles also the $k = 2$ and $n = 4$ case. BTW, where do you get all those questions? They are very interesting but I wonder what is the motivation for them. – levap Dec 26 '16 at 23:44

## 1 Answer

You have suggested the following procedure to construct an inner product on $\Lambda^k(H)$:

1. Choose some inner product $h \colon H \times H \rightarrow \mathbb{R}$ on $H$.
2. Choose some linear isomorphism $A \colon H \rightarrow H$.
3. Define an inner product $\gamma(h,A)$ on $\Lambda^k(H)$ by the formula $$\gamma(h,A)(v_1 \wedge \dots \wedge v_k, w_1 \wedge \dots \wedge w_k) := \det( h(Av_i, Aw_j))_{i,j=1}^k.$$

In your notation, $h = \left< \cdot, \cdot \right>$ and

$$\gamma(h,A)(v_1 \wedge \dots \wedge v_k, w_1 \wedge \dots \wedge w_k) = \left< \Lambda^k(A)(v_1 \wedge \dots \wedge v_k), \Lambda^k(A)(w_1 \wedge \dots \wedge w_k) \right> = \left< Av_1 \wedge \dots \wedge Av_k, Aw_1 \wedge \dots \wedge Aw_k \right> = \det( \left< Av_i, Aw_j \right>)_{i,j=1}^k.$$

Now, this procedure is extremely redundant in the following sense. You could have started with an arbitrary fixed inner product $g$ on $H$ and take a positive definite map $T \colon H \rightarrow H$ (with respect to $g$) and construct $\gamma(g,T)$. In fact, these already constructs all the metrics of the form $\gamma(h,A)$ for all $h,A$. Let me state this precisely and prove it. Fix some inner product $g$ on $H$.

Claim: Given $h,A$ there exists a $g$-positive definite map $T \colon H \rightarrow H$ such that $\gamma(h,A) = \gamma(g,T)$. Proof: Let $A = UP$ be the polar decomposition of $A$ with respect to $h$. Let $S$ be the map determined uniquely by $g(Sv,w) = h(Pv,Pw)$. Then $S$ is $g$-symmetric and positive. Finally, let $T = \sqrt{S}$ be the $g$-square root of $S$. We have

$$h(Av, Aw) = h(UPv, UPw) = h(Pv,Pw) = g(Sv,w) = g(T^2v,w) = g(Tv,Tw)$$

but then

$$\gamma(h,A)(v_1 \wedge \dots \wedge v_k, w_1 \wedge \dots \wedge w_k) = \det( h(Av_i, Aw_j))_{i,j=1}^k = \det (g(Tv_i, Tw_j))_{i,j=1}^k = \gamma(g,T)(v_1 \wedge \dots \wedge v_k, w_1 \wedge \dots \wedge w_k).$$

Thus, the image $\gamma(S,\operatorname{GL}(H))$ is in fact the same as $\gamma(g,\mathcal{P})$ where $\mathcal{P}$ denotes the set of $g$-positive definite operators on $H$. Thus, we have a map $\psi \colon \mathcal{P} \rightarrow \mathcal{T}$ given by $\psi(T) = \gamma(g,T)$ which has the same image as $\gamma$. Now, $\dim \mathcal{P} = \frac{n(n+1)}{2}$ while $\dim \mathcal{T} = \frac{{n \choose k} \left( {n \choose k} + 1 \right)}{2}$ so in general, the map $\psi$ cannot be surjective.

For example, when $n = 4$ and $k = 2$, we have $\dim \mathcal{P} = 6$ while $\dim \mathcal{T} = 21$.

• Alternatively, the image is the same as $\gamma(S,\operatorname{Id})$ which is even easier to show. – levap Dec 26 '16 at 23:51