# Hodge Star basis-independence proof

I am trying to solve a problem form a workbook I found, I apologize in advanced for the notations and my grammar, since english is not my first language I might have some grammatical errors and some notation differences in my question.

Te exercise states as follows:

Given $V$ oriented vector space, with internal product and a finite dimension n.

Fixing $\beta =\{e_1,...e_n\}$ orthonormal, positive oriented basis for V, and defining $e_i^{*}$ as $e_i$'s equivalent in the dual basis $\beta^*$. We define the hodge star as the linear operator $\star$: $\cup_p \Lambda^p(V^*) \to \cup_p \Lambda^p(V^*)$ that satisfies:

1. $\star(e_{i1}^{*}\land.....\land e_{ip}^{*})$ = $e_{j1}^{*}\land.....\land e_{jp}^{*}$ (Which means $\star(\Lambda^p(V^*)) \subset \Lambda^{n-p}(V^*)$)

2. $e_{i1}^{*}\land.....\land e_{ip}^{*}\land\star(e_{i1}^{*}\land.....\land e_{ip}^{*})= e_{1}^{*}\land.....\land e_{n}^{*}$

Prove that this definition of $\star$ does not depend on the basis chosen, as long as the basis is orthonormal and positive oriented.

Okay... so this si what I thought, out of the definition and the linearity of $\star$ you can calculate $\star(w)$ in fucntion of the dual basis. Using that $\star$ is linear all you got to know is $\star(e_{i1}^{*}\land.....\land e_{ip}^{*})$ which using the propperties above will be of the form:

$\star(e_{i1}^{*}\land.....\land e_{ip}^{*})$ = $(-1)^{\lambda}\land_{j \in \{i_1,..,i_n\}^c}e_{j}^{*}$ (with the subindex ordered, from lowest to highest and $\lambda$ a calculable interger).

So I thought next on grabbing another basis $\{v_1,...,v_n\}$ of $V^*$ and calculate $\star(v_{i1}\land....\land v_{ip})$ and see if the two propperties apply (Using that $v_{i1}\land....\land v_{ip}$ can be written as an unique linear combination of the vectors $\{e_{i1}^{*}\land.....\land e_{ip}^{*}: 1\leq i_1<...<i_p\leq n \}$).

My problem is I do not know where to apply the orthogonality and the fact that the other basis is positive, so I am stuck at this point.

Any help would be appreciated, thanks in advanced. Sorry again, for my grammar.

• HINT: If $T\colon\Bbb R^p\to\Bbb R^p$ is a linear map carrying one positively-oriented orthonormal basis to another, what is $\det(T)$? Jul 30 '17 at 17:26
• $det(T)$ should be positive. Maybe I am missing something but I still do not see where to apply $det$. Jul 30 '17 at 20:05
• Properties of $\wedge$ will give you exactly a determinant as the multiplicative factor when you write one basis in terms of the other. Indeed, on an oriented $n$-dimensional vector space (with an inner product), $v_1\wedge\dots\wedge v_n$ is precisely the signed volume of the parallelepiped, which is exactly the determinant of the matrix whose columns are the coordinates of those vectors with respect to an oriented, orthonormal basis. You need to work this out for starters. And the answer to my question is $1$. :) Jul 30 '17 at 20:08
• Ok.. I get it, did not take in count the fact that they are orthonormal vectors. But I think i get where to go from there. Thank you! Jul 30 '17 at 20:22
• You're welcome :) Jul 30 '17 at 20:22

First a example in dimension 3 with $(e_i)$ and $(f_i)$ two direct orthonormal basis. We define the star operator on $\Lambda^2(V^*)$ with the $(e_i)$ basis and try to state the same relation with the $(f_i)$ basis. For example in $\Lambda^2(V^*)$: $$\star(f_1^*\land f_2^*)=\star((a_1^1e_1^*+a_1^2e_2^*+a_1^3e_3^*)\land(a_2^1e_1^*+a_2^2e_2^*+a_2^3e_3^*))\\ =\star((a_1^1a_2^2-a_1^2a_2^1)e_1^*\land e_2^*+(a_1^1e2^3-a_1^3a_2^1)e_1^*\land e_3^*+(a_1^2e_2^3-a_1^3a_2^2)e_2^*\land e_3^*)\\ =(a_1^1a_2^2-a_1^2a_2^1)e_3^*-(a_1^1e_2^3-a_1^3a_2^1)e_2^*+(a_1^2e_2^3-a_1^3a_2^2)e_1^*$$ and we have to prove that this is equal to $$f_3^*=a_3^1e_1^*+a_3^2e_2^*+a_3^3e_3^*$$ So we must have $$\left \{ \begin{array}{c} a_1^1a_2^2-a_1^2a_2^1=a_3^1 \\ a_1^1e_2^3-a_1^3a_2^1=-a_3^2 \\ a_1^2e_2^3-a_1^3a_2^2=a_3^3 \end{array} \right.$$

This is a consequence of the relation $P=\operatorname{adj}(P)$ that comes from $P^tP=I_3$

For the general case the method is the same and the relation on orthogonal matrix is detail here.

• Thank you for your response, reading my question, I think my problem was that I did not take into account that the basis was orthonormal. Sorry that I cannot truly appreciate your answer since, I might need to refresh my multilinear forms to really understand what my problem was. May 31 '18 at 22:20

Perhaps someone can provide a more detailed answer, but I think I can sketch the idea. Essentially, the main defining characteristic of the Hodge star operator on a vector space is what you called condition 2.)

$$\star(e^{*}_{i_{1}} \wedge \cdots \wedge e^{*}_{i_{p}}) \wedge (e^{*}_{i_{1}} \wedge \cdots \wedge e^{*}_{i_{p}})=e^{*}_{i_{1}} \wedge \cdots \wedge e^{*}_{i_{n}}.$$

In words, this says that $\star A \wedge A$ is equal to the "volume form" of the vector space. This is more intuitive with a compact Riemannian manifold, but you can imagine that the "volume" of a vector space is just a scalar that can be normalized to whatever you like. So the point is that the righthand side, is simply a number, independent of the basis. Since the basis is orthogonal, if you chose a different basis, recall that the effect of such a transformation on the volume form is that you acquire a factor depending on the determinant. But by the orthogonality of the basis, this term should be unity and therefore, the Hodge star is independent of basis.

Another reason for the orthogonality, imagine $V$ were two dimensional and you chose non-orthogonal basis $(A,B)$. Then you can write $A$ as a component parallel to $B$ and one perpendicular to $B$. This would decompose the "volume form" $A \wedge B$ into two terms. And the one involving the component of $A$ parallel to $B$ would vanish by the anti-symmetry of the wedge product. So it suffices to consider orthogonal bases.