Let $V$ be an $n$-dimensional vector space with inner product $\langle,\rangle$ and volume element $\mathrm{vol} \in \mathrm{Alt}^n(V)$. Let $v \in \mathrm{Alt}^1(V)$ and $$F_v: \mathrm{Alt}^p(V) \to \mathrm{Alt}^{p+1}(V)$$ be the map $$F_v(\omega)=v\wedge \omega.$$ Show that the map $F^*_v=(-1)^{np} \star \circ F_v \circ \star : \mathrm{Alt}^{p+1}(V) \to \mathrm{Alt}^p(V)$ is adjoint to $F_v$ i.e. $\langle F_v\omega, \tau\rangle = \langle \omega, F^*_v\tau\rangle$.

I'm trying to show this for basis vectors like so. If we pick an orthonormal base $\{e_1, \dots, e_n\}$ for $V$ we get an orthonormal basis $\{\varepsilon_{i_1} \wedge \dots \wedge \varepsilon_{i_p} \mid 1 \le i_1 < \dots < i_p \le n\}$ for $\mathrm{Alt}^{p}(V)$. Now I think It would suffice to show that the equality $$\langle F_v(\varepsilon_{i_1} \wedge \dots \wedge \varepsilon_{i_p}), \varepsilon_{j_1} \wedge \dots \wedge \varepsilon_{j_p} \rangle = \langle \varepsilon_{i_1} \wedge \dots \wedge \varepsilon_{i_p}, F^*_v(\varepsilon_{j_1} \wedge \dots \wedge \varepsilon_{j_p})\rangle$$

holds for two basis vectors $\varepsilon_{i_1} \wedge \dots \wedge \varepsilon_{i_p}$ and $\varepsilon_{j_1} \wedge \dots \wedge \varepsilon_{j_p}$.

Looking at the left-hand side we have $$\begin{align*}\langle F_v(\varepsilon_{i_1} \wedge \dots \wedge \varepsilon_{i_p}), \varepsilon_{j_1} \wedge \dots \wedge \varepsilon_{j_p} \rangle &= \langle v \wedge \varepsilon_{i_1} \wedge \dots \wedge \varepsilon_{i_p}, \varepsilon_{j_1} \wedge \dots \wedge \varepsilon_{j_p} \rangle \end{align*}$$

but on the right-hand side I don't know what $F^*_v(\varepsilon_{j_1} \wedge \dots \wedge \varepsilon_{j_p})$ computes to. I have $$\begin{align*} F^*_v(\varepsilon_{j_1} \wedge \dots \wedge \varepsilon_{j_p}) &= (-1)^{np} \circ \star \circ F_v \circ \star(\varepsilon_{j_1} \wedge \dots \wedge \varepsilon_{j_p}) \end{align*}$$

but I don't have a formula for $\star(\varepsilon_{j_1} \wedge \dots \wedge \varepsilon_{j_p})$. What I do know is that if we instead use shuffles we have $$\star(\varepsilon_{\sigma(1)} \wedge \dots \wedge \varepsilon_{\sigma(p)}) = \mathrm{sign}(\sigma) \varepsilon_{\sigma(p+1)} \wedge \dots \wedge \varepsilon_{\sigma(n)}.$$ So this would give $$\begin{align*} F^*_v(\varepsilon_{\sigma(1)} \wedge \dots \wedge \varepsilon_{\sigma(p)}) &= (-1)^{np} \circ \star \circ F_v \circ \star(\varepsilon_{\sigma(1)} \wedge \dots \wedge \varepsilon_{\sigma(p)}) \\ &= (-1)^{np} \circ \star \circ F_v(\mathrm{sign}(\sigma) \varepsilon_{\sigma(p+1)} \wedge \dots \wedge \varepsilon_{\sigma(n)}) \\ &= (-1)^{np} \circ \star(v \wedge (\mathrm{sign}(\sigma) \varepsilon_{\sigma(p+1)} \wedge \dots \wedge \varepsilon_{\sigma(n)})) \\ &= (-1)^{np} \circ \star(v) \wedge \star(\mathrm{sign}(\sigma) \varepsilon_{\sigma(p+1)} \wedge \dots \wedge \varepsilon_{\sigma(n)})\end{align*}$$

but I'm unfortunatey stuck again as I don't know what $\star(v)$ and $\star(\mathrm{sign}(\sigma) \varepsilon_{\sigma(p+1)} \wedge \dots \wedge \varepsilon_{\sigma(n)})$ evaluates to.

If anyone knows that is this even the right approach I would appreciate the advice?

  • $\begingroup$ You don't need to use a basis to show this. Try writing out $\langle \omega, (-1)^{np}\ast F_v(\ast\tau)\rangle$ and simplifying. $\endgroup$ Commented Feb 18, 2023 at 14:15
  • $\begingroup$ @MichaelAlbanese It's the same issue with this. I end up in trouble with the computation of $(-1)^{np}\ast F_v(\ast\tau)$. What I thought was that it would have been easier with basis elements, but I was wrong. $\endgroup$
    – Homer
    Commented Feb 18, 2023 at 14:57
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    $\begingroup$ But you're not taking the Hodge star of $(-1)^{np}$, you're taking the Hodge star of $(-1)^{np}$ multiplied by a form. Your sign is incorrect. It is not true that $\omega\wedge v = -v\wedge\omega$ in general. $\endgroup$ Commented Feb 18, 2023 at 17:41
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    $\begingroup$ Also, $\ast\ast = (-1)^{p(n-p)}$ on $p$-forms, but $F_v(\ast\tau)$ is not a $p$-form, so your sign there is also off. $\endgroup$ Commented Feb 18, 2023 at 17:43
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    $\begingroup$ You're correct, of course, my bad. Keep in mind that $(-1)^{2np-p^2} = (-1)^{-p^2} = (-1)^p$ and $\omega\wedge v = (-1)^{1\times p}v\wedge\omega = (-1)^pv\wedge\omega$. Putting it all together, you should be able to finish it off. $\endgroup$ Commented Feb 18, 2023 at 18:48

1 Answer 1


Let me summarise the discussion in the comments regarding a basis free proof.

The map $\mathbb{R} \to \operatorname{Alt}^nV$ given by $r \mapsto r\operatorname{vol}$ is an isomorphism, so it is enough to show that $\langle F_v(\omega), \tau\rangle\operatorname{vol} = \langle\omega, F_v^*(\tau)\rangle\operatorname{vol}$. Using the fact that $\langle\alpha,\beta\rangle\operatorname{vol} = \alpha\wedge\ast\beta$, we have

\begin{align*} & \langle\omega, F_v^*(\tau)\rangle\operatorname{vol}\\ =&\ \omega\wedge\ast F_v^*(\tau)\\ =&\ \omega\wedge\ast (-1)^{np}\ast F_v(\ast\tau)\\ =&\ \omega\wedge (-1)^{np}\ast\ast F_v(\ast\tau) && (\ast\ \text{is linear})\\ =&\ \omega\wedge (-1)^{np}(-1)^{(n-p)p} F_v(\ast\tau) &&(\ast\ast\ \text{is multiplication by}\ (-1)^{k(n-k)}\ \text{on}\ \operatorname{Alt}^k(V))\\ =&\ \omega\wedge(-1)^{np + np - p^2}F_v(\ast\tau)\\ =&\ \omega\wedge(-1)^pv\wedge\ast\tau &&((-1)^{-p^2} = (-1)^p)\\ =&\ v\wedge\omega\wedge\ast\tau && (\alpha\wedge\beta = (-1)^{ab}\beta\wedge\alpha\ \text{if}\ \alpha\in \operatorname{Alt}^a(V), \beta\in\operatorname{Alt}^b(V))\\ =&\ F_v(\omega)\wedge\ast\tau\\ =&\ \langle F_v(\omega), \tau\rangle\operatorname{vol}. \end{align*}


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