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 $\{\varepsilon^i\}$ ?

  • $\begingroup$ I don't get it. I mean, if $V$ is $2$-dimensional, then most exterior powers of $V$ are trivial, right? $\endgroup$ Feb 16, 2016 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, 2016 at 16:10

1 Answer 1


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.


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*}


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .