# Invariant proof for a commutation identity involving the Hodge stars on $V,V^*$

Let $V$ be a $d$-dimensional oriented inner product space, and let $\alpha \in V^*$.

Let $\star_V,\star_{V^*}$ denote the corresponding Hodge star operators on $V,V^*$ (of course, I take the inner product and orientaion on $V^*$ to be the one induced on it by those of $V$).

I am trying to prove the following identity:

$$(1) \, \, \, (\star_{V^*}^1 \alpha)(v_1,...,v_{d-1})=(-1)^{d-1} \alpha(\star_V^{d-1}(v_1 \wedge \dots \wedge v_{d-1}))$$ for every $v_1,...,v_{d-1} \in V$. In some sense, this identity asserts that Hodge-stars and linear functionals commute.

Note that $\star_{V^*}^1 \alpha \in \Lambda_{d-1}(V^*) \cong A_{d-1}(V)$ is an alternating map of degree $d-1$, so indeed both sides of the equation are real numbers.

I succeeded to prove this after picking an orthonormal basis for $V$ (see details below).

Question: Is there an invariant proof, which uses only the abstract definition of the Hodge stars, without refering to bases and "coordinates" formulas. (i.e a proof which relies solely on the characterization of $*$ via $(v,w)=*(w \wedge *v)$)

My proof:

Let $v_i$ be a positively oriented orthonormal basis of $V$. It is enough to prove equality $(1)$ for $v_1,...v_{d-1}$.

Since $\star_V^{d-1}(v_1 \wedge \dots \wedge v_{d-1}) =v_d$, the equality redueces to:

$$(1') \, \, \, (\star_{V^*}^1 \alpha)(v_1,...,v_{d-1})=(-1)^{d-1}\alpha(v_d)$$

Denote by $v^i$ the dual basis of $v_i$. Write $\alpha$ in components:

$$\alpha=\alpha_jv^j$$

Since $v^j$ is a positive orthonormal basis for $V^*$, we get $$(\star_{V^*}^1 v^j)=(-1)^{j-1} v^1 \wedge \dots \hat v^j \dots \wedge v^d$$ where the $\hat {}$ denotes elements which are being omitted.

Thus, $$(\star_{V^*}^1 v^j)(v_1,...,v_{d-1})=(-1)^{j-1} v^1 \wedge \dots \hat v^j \dots \wedge v^d (v_1,...,v_{d-1})= \left\{ \begin{array}{ll} 0 & \mbox{if } j \neq d \\ (-1)^{d-1} & \mbox{if } j = d \end{array} \right.$$

Finally, by the linearity of the Hodge-star operator we obtain

$$(\star_{V^*}^1 \alpha)(v_1,...,v_{d-1})=\alpha_j \big( (\star_{V^*}^1v^j)(v_1,...,v_{d-1}) \big)=(-1)^{d-1}\alpha_d=(-1)^{d-1}\alpha(v_d)$$

It seems that you have a sign error. The one before last line in your calculation should end with $(-1)^{d-1}$ unless you use a strange sign convention for the pairing $\Lambda^k(V) \times \Lambda^k(V^{*}) \rightarrow \mathbb{R}$.

I can offer an alternative point of view and argument that doesn't use coordinates directly but does use another property of the Hodge star which I don't see how one can prove without introducing a basis. In what follows, all the vector spaces involved are real and finite dimensional. To fix notation, I will denote by $\Lambda(V)$ the exterior algebra on $V$ and use the pairing $\Lambda^k(V) \times \Lambda^k(V^{*}) \rightarrow \mathbb{R}$ given by

$$(v_1 \wedge \dots \wedge v_k , \varphi^1 \wedge \dots \wedge \varphi^k) = \det(\varphi^i(v_j))$$ in order to identify elements of $\Lambda^k(V^{*})$ as functionals on $\Lambda^k(V)$.

1. Given an inner product on $\left< \cdot, \cdot \right>_V$ on $V$, it induces naturally an inner product $\left< \cdot, \cdot \right>_{\Lambda(V)}$ on $\Lambda(V)$. This construction is functorial in the sense that if $T \colon (V, \left< \cdot, \cdot \right>_V) \rightarrow (W, \left< \cdot, \cdot \right>_W)$ is an isometry then $\Lambda(T) \colon (\Lambda(V), \left< \cdot, \cdot \right>_{\Lambda(V)}) \rightarrow (\Lambda(W), \left< \cdot, \cdot \right>_{\Lambda(W)})$ is also an isometry of inner product spaces. This can be verified without choosing bases.
2. Given an inner product on $g = \left< \cdot, \cdot \right>_V$ on $V$ and an orientation $\omega \in \Lambda^{\text{top}}(V)$ that is compatible with the inner product in the sense that $\left< \omega, \omega \right> = 1$, we can define a linear operator $\star_{(V,g,\omega)} = \star_V \colon \Lambda(V) \rightarrow \Lambda(V)$ that is determined uniquely by the following two properties:

1. The operator $\star_V$ maps $\Lambda^k(V)$ into $\Lambda^{n-k}(V)$ for all $0 \leq k \leq n = \dim V$.
2. For all $0 \leq k \leq \dim V$ and $\alpha, \beta \in \Lambda^k(V)$ we have $$\alpha \wedge \star_V (\beta) = \left< \alpha, \beta \right>_{\Lambda^k(V)} \omega.$$

The two properties above determine $\star_V$ uniquely since the pairing $\Lambda^k(V) \times \Lambda^{n-k}(V) \rightarrow \mathbb{R}$ induced by the choice of orientation is non-degenerate. This involves some argument that uses the structure of $\Lambda^k(V)$ and is best seen by choosing a basis.

3. The Hodge star is an isomorphism. This follows from the defining property as $\alpha \wedge \star \alpha = \left< \alpha, \alpha \right>_{\Lambda^k(V)} \omega$ and so if $0 \neq \alpha \in \Lambda^k(V)$ and $\star \alpha = 0$ then $\left< \alpha, \alpha \right>_{\Lambda^k(V} = 0$ and since $\star$ behaves well with respect to the grading and the dimensions are right (I guess this uses a basis argument in the background), the result follows.


5. The Hodge star is an isometry. This is usually proved by analyzing explicitly the Hodge star action on an orthonormal basis and unfortunately I don't see how one can avoid here an argument that is more or less comparable to choosing an orthonormal basis. The point is that any proof much use the explicit form of the inner product on $\Lambda(V)$ which is defined on elementary $k$-wedges using the inner product on $V$ and then extended linearly. The abstract characterizing property doesn't even make it clear that one can choose a basis of $\Lambda^k(V)$ of elementary $k$-wedges that $\star$ sends to elementary $n-k$-wedges and without this information, not much more can be deduced.

6. If $\dim V = n$ and $\alpha \in \Lambda^k(V)$ then $(\star_V \circ \star_V)(\alpha) = (-1)^{k(n-k)} \alpha$. This follows from (and is actually equivalent to) the previous item and the defining property as $$(\star \beta) \wedge (\star (\star \alpha)) = \left< \star \beta, \star \alpha \right> \omega = \left< \alpha, \beta \right> \omega = \alpha \wedge \star \beta = (\star \beta) \wedge ((-1)^{k(n-k)} \alpha)$$ for all $\beta \in \Lambda^k(V)$ and since $\star$ is an isomorphism and the pairing is non-degenerate, we get the required result.

Finally, let us use the properties above to prove the result. Denote by $T \colon V \rightarrow V^{*}$ the isometry obtained using the inner product on $V$ (so $T(v) = \left< v, \cdot \right>$). Note that if $$u_1 \wedge \dots \wedge u_k, v_1 \wedge \dots \wedge v_k \in \Lambda^k(V)$$ we have $$(\Lambda(T)(u_1 \wedge \dots \wedge u_k))(v_1 \wedge \dots \wedge v_k) = (Tu_1 \wedge \dots \wedge Tu_k)(v_1 \wedge \dots \wedge v_k) \\ = \det(T(u_i)(v_j)) = \det( \left< u_i, v_j \right> ) = \left< u_1 \wedge \dots \wedge u_k, v_1 \wedge \dots \wedge v_k \right>.$$

By bilinearity, we get that $(\Lambda(T)(\alpha))(\beta) = \left< \alpha, \beta \right>$ for all $\alpha, \beta \in \Lambda^k(T)$.

Now let $\varphi \in \Lambda^k(V^{*})$ and choose $\alpha \in \Lambda^k(V)$ such that $\Lambda(T)(\alpha) = \varphi$. By the naturality of the Hodge star, we get for all $\beta \in \Lambda^{n-k}(V)$

$$(\star_{V^*} \varphi)(\beta) = ((\star_{V^{*}} \circ \Lambda(T))(\alpha))(\beta) = ((\Lambda(T) \circ \star_V)(\alpha))(\beta) = (\Lambda(T)(\star_V(\alpha)))(\beta) = \left< \star_V \alpha, \beta \right> = (-1)^{k(n-k)} \left< \alpha, \star_V \beta \right> = (-1)^{k(n-k)}(\Lambda(T)(\alpha))(\star_V \beta) = (-1)^{k(n-k)}\varphi(\star_V \beta).$$

Your result is obtained by taking $k = 1$ and $\beta = v_1 \wedge \dots \wedge v_{n-1}$ (with $\varphi$ instead of $\alpha$ and $n$ instead of $d$).

Addendum: There is an argument that proves that the Hodge star is an isometry that doesn't involve analyzing the action of $\star$ on an orthonormal basis and is more coordinate free but requires much more work. Let $(U,g,\omega)$ and $(V,h,\nu)$ be two finite dimensional oriented inner product spaces and consider the direct sum $(U \oplus V, g \oplus h)$. I will continue to denote by $g$ the inner product induced by $g$ on $\Lambda(U)$ and similarly for $V$

Lemma: There is a natural isometry $\varphi \colon (\Lambda(U) \otimes \Lambda(V), g \otimes h) \rightarrow (\Lambda(U \oplus V), g \oplus h)$ of bi-graded graded-commutative algebras endowed with an inner product. The product structure on the tensor product $\Lambda(U) \otimes \Lambda(V)$ is defined by the usual formula for the product of graded-commutative algebras: $$(\alpha \otimes \beta) \hat{\wedge} (\gamma \otimes \delta) := (-1)^{\deg \beta \deg \gamma} (\alpha \wedge \gamma) \otimes (\beta \wedge \delta).$$ The isomorphism is determined by its action on elementary tensors by the formula $$\varphi((u_1 \wedge \dots \wedge u_i) \otimes (v_1 \wedge \dots \wedge v_j)) = u_1 \wedge \dots \wedge u_i \wedge w_1 \wedge \dots \wedge w_j.$$

The proof is a tedious verification that everything makes sense, is well-defined and behaves as expected. The reason we get an isometry is that $U$ and $V$ are orthogonal inside $U \oplus V$ and so when we compute the induced inner product, the terms that mix vectors from $U$ and $V$ die.

In particular, using $\varphi$, we can define a volume form on $U \oplus V$ by $\varphi(\omega \otimes \nu)$ (this is the standard induced volume form on a direct sum) and consider the Hodge star of $(\Lambda(U \oplus V), g \oplus h, \varphi(\omega \otimes \nu))$.


Proof: Let $\alpha, \gamma \in \Lambda(U)$ and $\beta, \delta \in \Lambda(V)$ be homogeneous elements. We calculate $$\varphi(\alpha \otimes \beta) \wedge \star(\varphi(\gamma \otimes \delta)) = \left< \varphi(\alpha \otimes \beta), \varphi(\gamma \otimes \delta) \right>_{g \oplus h} \varphi(\omega \otimes \nu) = \varphi( \left< \alpha \otimes \beta, \gamma \otimes \delta \right>_{g \otimes h} \omega \otimes \nu) = \varphi( (\left< \alpha, \gamma \right>_g \omega) \otimes (\left< \beta, \delta \right>_h \nu)) = \varphi( (\alpha \wedge \star_U \gamma) \otimes (\beta \wedge \star_V \delta)) = (-1)^{\deg \beta \deg \star_U \gamma} \varphi( (\alpha \otimes \beta) \hat{\wedge} (\star_U \gamma \otimes \star_V \delta)) \\ = (-1)^{\deg \beta \deg \star_U \gamma} \varphi(\alpha \otimes \beta) \wedge (\varphi \circ (\star_U \otimes \star_V))(\gamma \otimes \delta).$$

By taking $\alpha, \beta$ with $\deg \alpha = \deg \gamma$ and $\deg \beta = \deg \delta$ and using the non-degeneracy of the (bi)-pairing, we see that the diagram commutes with a sign factor of

$$\star\varphi(\gamma \otimes \delta) = (-1)^{(\dim U - \deg \gamma)\deg \delta} \varphi(\star \gamma \otimes \star \delta).$$

Of course, an alternative proof would involve choosing orthonormal bases for $U,V$ and analyzing the operators explicitly. A less abstract way of stating the proposition above is that if $\gamma \in \Lambda(U)$ and $\delta \in \Lambda(V)$ and $U \perp V$ then

$$\star(\gamma \wedge \delta) = (-1)^{(\dim U - \deg \gamma) \deg \delta} (\star \gamma) \wedge (\star \delta).$$

Finally, we can prove that $\star$ is an isometry by an inductive argument. Almost by definition, the Hodge star on an $n$-dimensional vector space is an isometry between $\Lambda^n(V)$ and $\Lambda^0(V)$ and in particular it is an isometry on the whole of $\Lambda(V)$ if $\dim V \leq 1$. If $\dim V > 1$, split it into a direct sum of orthogonal subspaces of lower dimension and use the diagram above and the induction hypothesis. The specific sign factor plays no role in showing that $\star$ is an isometry.

• BTW, if you take $V = W$ in $4$, you get that $\star_V$ commutes with all the operators of the form $\Lambda(T)$ where $T \in SO(V, \left< \cdot, \cdot \right>)$. Namely, $\star_V$ is an automorphism of the $SO(V, \left< \cdot, \cdot \right>)$ representation $\Lambda(V)$. – levap Oct 27 '16 at 18:59
• Thanks! You took the words right out of my mouth... I was thinking lately about the equivalence between the two facts you mentioned;(1) $\star$ is an isometry , and (2) $\star \star=(-1)^{k(n-k)}$. I indeed saw they were equivalent, but could not prove either one of them "from scratch" without choosing a basis. I guess your comment at item $5$ makes sense and there is probably no way to avoid this entirely... – Asaf Shachar Oct 31 '16 at 16:10
• @AsafShachar: You're welcome. I actually thought about it some more and managed to come up with an argument that doesn't use explicitly a choice of an orthonormal basis and gives you more information but in all other senses is much more convoluted. See how much you like it. – levap Oct 31 '16 at 22:15
• Thanks. Your new proof is interesting, and has some "natural" aura. However, I agree with you it's much too complicated.... I guess sometimes doing things more invariantly doesn't make them simpler. Againg, on the whole I am quite happy with the result. It is really not so bad to prove either of the two mentioned properties of the Hodge star via bases. (In fact the proofs are almost trivial). – Asaf Shachar Nov 1 '16 at 17:14
• The main point in my opinion, is not to work with bases every time we prove some "small" property, but to try to use (like you did) the few basic properties of $\star$. Indeed, your proof (and reformulation/generalization of the claim) made the statement look much more natural. – Asaf Shachar Nov 1 '16 at 17:14