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Frank Warner's book, chapter 2, excercise 13 states the following:

If $V$ is an oriented inner product space ($n$ dimensional) there is a linear map $\ast \colon \Lambda (V) \to \Lambda (V)$, called star, which is well-defined by the requirement that for any orthonormal basis $e_1,\dots,e_n$ of $V$ (in particular, for any re-ordering of a given basis), $\ast(1) = \pm e_1 \wedge \dots \wedge e_n$, $\ast(e_1 \wedge \dots \wedge e_n) = \pm 1$, $\ast(e_1 \wedge \dots \wedge e_k) = \pm e_{k+1} \wedge \dots \wedge e_n$, where one takes "+" if $e_1 \wedge \dots \wedge e_n$ lies on the component of $\Lambda^n(V)-\{0\}$ determined by the orientation and "-" otherwise. Observe that $\ast\colon \Lambda^k(V) \to \Lambda^{n-k}(V)$. Prove that on $\Lambda^k(V)$, $\ast\ast = (-1)^{k(n-k)}$.

A prior part of that exercise was to show that if we have an ONB $e_1,\dots,e_n$ of $V$ then the corresponding basis of $\Lambda(V)$ is orthonormal (after extending the inner product to $\Lambda(V)$ in the usual way), which I already accomplished.

But now I'm having trouble with his definition of $\ast$. How can I calculate $\ast$ on all other basis elements, i.e. $\ast(e_{i_1}\wedge \dots \wedge e_{i_k})$? What exactly does he mean with "for any re-ordering of a given basis"? Or is there any other way to prove the claim?

Clarification: I know how the $\ast$ acts on general elements from many other sources, my question here is, if the claim follows by only knowing $\ast(e_1 \wedge \dots \wedge e_k) = e_{k+1} \wedge \dots \wedge e_n$ if my basis has the "right" orientation.

Edit: For a solution see below.

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Given an arbitrary $k$-element ordered list $i_1,i_2,\dots,i_k$ (consisting of distinct elements), choose complementary numbers $j_{k+1},\dots,j_n$ so that $\{i_1,\dots,i_k,j_{k+1},\dots,j_n\} = \{1,2,\dots,n\}$. If you order the $j$'s so that $$e_{i_1}\wedge e_{i_2} \wedge \dots \wedge e_{i_k}\wedge e_{j_{k+1}}\wedge \dots \wedge e_{j_n} = e_1\wedge e_2\wedge\dots\wedge e_n,$$ then $\star(e_{i_1}\wedge e_{i_2} \wedge \dots \wedge e_{i_k}) = e_{j_{k+1}}\wedge \dots \wedge e_{j_n}$. In more concise notation, $\star e_I = \pm e_{I'}$, where $I'$ consists of the complementary indices to $I$, and the sign is chosen so that $e_I\wedge\star e_I = e_1\wedge\dots\wedge e_n$.

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  • $\begingroup$ Thanks for your answer! Maybe I was not clear enough, but my question is if that already follows from the fact that only $\ast(e_1 \wedge \dots \wedge e_k) = e_{k+1} \wedge \dots \wedge e_n$? The exercise in the book does not give more. $\endgroup$ – T'x Apr 22 '16 at 16:23
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    $\begingroup$ I'm thinking of picking a standard orthonormal basis to start with. Note that Warner specifically says any orthonormal basis, so you can reorder with your indices $I,I'$ and call that your orthonormal basis. Notice that he even says that in his definition. With his "any," you probably should check well-definedness. It's just a matter of keeping track of signs and of knowing that for any two orthonormal bases $e_1,\dots,e_n$ and $f_1,\dots,f_n$, we have $e_1\wedge\dots\wedge e_n = \pm f_1\wedge\dots\wedge f_n$ (why?). $\endgroup$ – Ted Shifrin Apr 22 '16 at 16:33
  • $\begingroup$ Ah, now the things are clearer, thank you! So I start with one ONB and apply his definition. Then by re-ordering (where sometimes the orientation changes - in that case we have a minus in front of the exterior product over the new order), I get the remaining images of basis elements of the exterior algebra. $\endgroup$ – T'x Apr 22 '16 at 16:42
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Since it might be of interest in the future, here's my solution only using Warner's definition $\star(e_1 \wedge \dots \wedge e_k) = e_{k+1} \wedge \dots \wedge e_n$ on a ONB $e_1,\dots,e_n$ with orientation determined by $e_1 \wedge \dots \wedge e_n$: We can exclude the easy cases $k=0$ and $k=n$.

We have $e_1 \wedge \dots \wedge e_n = \operatorname{sgn}(\pi) e_{\pi(1)} \wedge \dots \wedge e_{\pi(n)}$, so if we use a re-ordered basis, orientation changes by $\operatorname{sgn(\pi)}$.

Using Warner's definition for the re-ordered basis gives: $\star(e_{\pi(1)} \wedge \dots \wedge e_{\pi(k)}) = \operatorname{sgn}(\pi) e_{\pi(k+1)} \wedge \dots \wedge e_{\pi(n)}$.
Now we have $\star\star(e_{\pi(1)} \wedge \dots \wedge e_{\pi(k)}) = \operatorname{sgn}(\pi) \star( e_{\pi(k+1)} \wedge \dots \wedge e_{\pi(n)})$.

To use Warner's definition again we have to re-order the basis again and calculate how the orientation changes: $e_{\pi(k+1)} \wedge \dots \wedge e_{\pi(n)} \wedge e_{\pi(1)} \wedge \dots \wedge e_{\pi(k)} = (-1)^{k(n-k)} e_{\pi(1)} \wedge \dots \wedge e_{\pi(n)}$.
Hence $\operatorname{sgn}(\pi) \star(e_{\pi(k+1)} \wedge \dots \wedge e_{\pi(n)}) = \operatorname{sgn}(\pi)^2 (-1)^{k(n-k)} e_{\pi(1)} \wedge \dots \wedge e_{\pi(k)}$.

This proves $\star\star = (-1)^{k(n-k)}$ since $\star$ is extended linearly.

Thanks to Ted Shifrin, who brought me on the right track.

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