5
$\begingroup$

Let $V$ be a $n$-dimensional vector space. Given an inner product on $V$ one may define an inner product on the simple $k$-vectors of $\Lambda^k(V)$ by $$\langle v_1 \wedge \cdots \wedge v_k, w_1 \wedge \cdots \wedge w_k\rangle_{\Lambda^k(V)} := \operatorname{det}\left(\langle v_i, w_j \rangle_V\right)$$ and extend it bilinearly. As usual this induces a norm on $\Lambda^k(V)$.

Burago/Ivanov claim in [Lemma 2.4, p. 6] that an oriented volume form $\omega\in \Lambda^2(V^\ast) \cong \left(\Lambda^2(V)\right)^\ast$ on $V$ determines a linear isometry $J: V \to V^\ast$ in "a standard way".

I don't understand the "isometry"-part. Here is what I have so far:

Define the mapping $J:V \to V^\ast$ by $J(u)(v) := \omega(u \wedge v), v\in V$. I can show that this is an isomorphism. I can define a somewhat "dual volume form" $\omega^\ast \in \Lambda^2(V) \cong \Lambda^2(V^{\ast\ast}) \cong \left(\Lambda^2(V^\ast)\right)^\ast$ by means of $$\omega^\ast(l\wedge g) := \omega\left(J^{-1}(l)\wedge J^{-1}(g)\right)$$ Thus, $$\omega^\ast\left(J(u)\wedge J(u')\right) = \omega(u\wedge u').$$ This looks quite promising already. (I am able to generalise this to $\Lambda^n(V)$ via $\widetilde J: \Lambda^{n-1}(V) \to V^\ast, \sigma \mapsto \omega(\sigma \wedge \cdot)$ and Hodge dual)

The way Burago/Ivanov use the "isometry"-part in their paper is $\left|J(v) \wedge J(v')\right| = \left|v \wedge v'\right|$ though (where the norms are on the respective exterior powers).

Is there a relationship between the induced norm on the exterior power and a corresponding volume form? Maybe by choosing an orthonormal basis for $V$ and taking the standard volume form $\varepsilon^1 \wedge \cdots \wedge \varepsilon^n$ determined by the dual basis $\{\varepsilon^i\}$ ?

$\endgroup$
2
  • $\begingroup$ I don't get it. I mean, if $V$ is $2$-dimensional, then most exterior powers of $V$ are trivial, right? $\endgroup$ Feb 16 '16 at 13:48
  • $\begingroup$ Yes. The only non-trivial ones are $V=\Lambda^1(V)$ and $\Lambda^2(V)$ where the latter is again isomorphic to $V$ by the Hodge dual. But does this make my question trivial? That would be great because as you can see below it took me a bit to prove the stated question $\endgroup$ Feb 16 '16 at 16:10
3
$\begingroup$

I will answer my own question in some more generality.

Let $V$ be an $n$-dimensional real inner product space. On $\Lambda^k(V)$ we define the inner product \begin{align} \left\langle v_1 \wedge \cdots \wedge v_k, w_1 \wedge \cdots \wedge w_k\right\rangle_{\Lambda^k(V)} := \operatorname{det}\left(\langle v_i, w_j \rangle_V\right) \end{align} and extend it bilinearly. This induces a norm $ \left\|\sigma\right\|_{\Lambda^k(V)} := \sqrt{\left\langle \sigma, \sigma\right\rangle_{\Lambda^k(V)}}. $

Given an orthonormal basis $\{e_1,e_2,\ldots,e_n\}$ for $V$ one can easily calculate that $e_1\wedge e_2\wedge \cdots\wedge e_n$ is a unit $n$-vector in the one-dimensional real vector space $\Lambda^n(V)$.

Introduce the basis $\{\varepsilon^1,\varepsilon^2,\ldots,\varepsilon^n\}$ for $V^\ast$ dual to $\{e_1,e_2,\ldots,e_n\}$, that is, $\varepsilon^i(e_j) = \delta^i_j$. The Riesz representation theorem gives an inner product on $V^\ast$ and the dual basis is orthonormal with respect to this inner product. Then analoguously, $\varepsilon^1\wedge \varepsilon^2\wedge \ldots\wedge \varepsilon^n$ is a unit $n$-form in $\Lambda^n(V^\ast)$.

Now let $\omega\in \Lambda^n(V^\ast)$ be the standard volume form given by $\omega = \varepsilon^1\wedge \varepsilon^2\wedge \ldots\wedge \varepsilon^n$. Then $\omega(e_1\wedge e_2\wedge \cdots\wedge e_n)=1$.

Any $n$-vector $v_1\wedge v_2\wedge \cdots\wedge v_n\Lambda^n(V)$ is a multiple $c$ of the basis $n$-vector $e_1\wedge e_2\wedge \cdots\wedge e_n$.

Lemma 1

If $v_1\wedge v_2\wedge \cdots\wedge v_n = c e_1\wedge e_2\wedge \cdots\wedge e_n$ then \begin{align} \left\|v_1\wedge v_2\wedge \cdots\wedge v_n\right\|_{\Lambda^n(V)} = \left|\omega(v_1\wedge v_2\wedge \cdots\wedge v_n)\right|. \end{align}

Proof: \begin{align} \left\|v_1\wedge v_2\wedge \cdots\wedge v_n\right\|_{\Lambda^n(V)} &= \frac{\left\|v_1\wedge v_2\wedge \cdots\wedge v_n\right\|_{\Lambda^n(V)}} {\left\|e_1\wedge e_2\wedge \cdots\wedge e_n\right\|_{\Lambda^n(V)}}\\ &= |c| \\ &= \left|\frac{\omega(v_1\wedge v_2\wedge \cdots\wedge v_n)} {\omega(e_1\wedge e_2\wedge \cdots\wedge e_n)}\right| \\ &= \left|\omega(v_1\wedge v_2\wedge \cdots\wedge v_n)\right| \end{align}

Let us define the map $\widetilde J: \Lambda^{n-1}(V) \to V^\ast$ by $\widetilde J(\sigma)(v) := \omega(\sigma \wedge v)$. This map is an isomorphism which can be seen by a straight forward calculation and I won't do it here.

The Hodge dual gives an isomorphism $\star: \Lambda^k(V) \to \Lambda^{n-k}(V)$ which is characterised by \begin{align} (\star \lambda) \wedge \theta = \left\langle\lambda,\theta\right\rangle_{\Lambda^{k}(V)} e_1\wedge e_2\wedge \cdots\wedge e_n \end{align} where $\lambda, \theta \in \Lambda^k(V)$ are arbitrary.

The composite map $J := \widetilde J\circ\star: V \to V^\ast$ is therefore also an isomorphism.

Define an $n$-vector $\omega^\ast := J_\ast \omega \in \Lambda^n(V) \cong \left(\Lambda^n(V^\ast)\right)^\ast$ as the pushforward of the volume form $\omega$ by $J$, that is, \begin{align} \omega^\ast(\phi^1\wedge \phi^2\wedge\cdots\wedge \phi^n) := \omega\left(J^{-1}(\phi^1)\wedge J^{-1}(\phi^2)\wedge\cdots\wedge J^{-1}(\phi^n)\right). \end{align} for $\phi^1\wedge \phi^2\wedge\cdots\wedge \phi^n\in \Lambda^n(V^\ast)$. This is well-defined because $J$ is an isomorphism and nonzero because $\omega$ is. Therefore, $\omega^\ast$ is a volume form on $V^\ast$ (what I called dual volume form in the question)

Lemma 2

For all $i=1,2,\ldots,n$ it holds that $J(e_i) = \epsilon^i$ and thus \begin{align*} \omega^\ast(\epsilon^1\wedge\epsilon^2\wedge\cdots\wedge\epsilon^n) &= 1. \end{align*}

Proof: By the equation for the Hodge dual we have \begin{align*} J(e_i)(e_j) &= (\widetilde J \circ \star)(e_i)(e_j) = \widetilde J (\star e_i)(e_j) \\ &= \omega(\star e_i \wedge e_j) \\ &= \omega\left(\left\langle e_i,e_j\right\rangle_{V} e_1\wedge e_2\wedge\cdots\wedge e_n\right) \\ &= \delta^i_j \omega\left(e_1\wedge e_2\wedge\cdots\wedge e_n\right) = \delta^i_j \end{align*} The second assertation follows from the definition of $\omega^\ast$.

Finally, we can prove the following result.

Proposition

Let $\{e_1,e_2,\ldots,e_n\}$ be an orthonormal basis for $V$ and $\omega$ the standard volume form. The isomorphism $J$ as above is isometric in the following sense \begin{align*} \left\|J(v_1)\wedge J(v_2)\wedge\cdots\wedge J(v_n)\right\|_{\Lambda^n(V^\ast)} = \left\|v_1\wedge v_2\wedge\cdots\wedge v_n\right\|_{\Lambda^n(V)} \end{align*} for all $v_1\wedge v_2\wedge\cdots\wedge v_n\in \Lambda^n(V)$.

Proof: Lemma 1 and 2 yield \begin{align*} \left\|J(v_1)\wedge J(v_2)\wedge\cdots\wedge J(v_n)\right\|_{\Lambda^n(V^\ast)} &= \frac{\left\|J(v_1)\wedge J(v_2)\wedge\cdots\wedge J(v_n)\right\|_{\Lambda^n(V^\ast)}} {\left\|\epsilon^1\wedge\epsilon^2\wedge\cdots\wedge \epsilon^n\right\|_{\Lambda^n(V^\ast)}} \\ &= \left|c\right| \\ &= \left|\frac{\omega^\ast\left(J(v_1)\wedge J(v_2)\wedge\cdots\wedge J(v_n)\right)} {\omega^\ast\left(\epsilon^1\wedge\epsilon^2\wedge\cdots\wedge \epsilon^n\right)}\right| \\ &= \left|\omega^\ast\left(J(v_1)\wedge J(v_2)\wedge\cdots\wedge J(v_n)\right)\right| \\ &= \left|\omega(v_1\wedge v_2\wedge \cdots \wedge v_n)\right| \\ &= \left\|v_1\wedge v_2\wedge \cdots \wedge v_n\right\|_{\Lambda^n(V)}. \end{align*}

$\endgroup$

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.