I'm very confused about how to approach this question. The question is

If $\omega\in\Lambda^n(V)$ is the volume element determined by $T$ and $\mu$, and $w_1,\dots,w_n\in V$, show that $$|\omega(w_1,\dots,w_n)|=\sqrt{\det(g_{ij})}$$ where $g_{ij}=T(w_i,w_j)$.

Hint: If $v_1,\dots,v_n$ is an orthonormal basis and $w_i=\sum_{j=1}^{n}a_{ij}v_j$, show that $g_{ij}=\sum_{k=1}^{n}a_{ik}a_{kj}$.

Now there are two parts to my confusion.

The first part is, by definition of the inner product, isn't the most general form of $T(w_i,w_j)=\sum_{k=1}^{n}a_{ik}a_{jk}$, essentially $Av\cdot A^Tv$, so the order of the indices would be opposite of what he gives?

The second part is, how does that inner product factor into finding the size of the volume element? I've considered taking the tensor product of the volume element with itself, to try and use the definition $|x|=\sqrt{x^2}$, but this seems to be a dead end. My best guess to how this factors into the proof is that he then uses Gram-Schmidt to produce an orthonormal basis using $w_i$, but this still doesn't explain how the square root gets there.

Could I get a hint towards how to solve this problem (I still want to give this a go, I just think I'm going in the wrong direction)? I think I've exhausted all of my ideas, but none of them seem to actually use the algebraic ideas w.r.t. tensors that he has brought up in the chapter.

Edit: Some other things I noticed about the question are that the inner product only seems to hold if $A$ (for $w_i=Av$) is symmetric ($A=A^T$), which would essentially mean $|\omega(w_1,\dots,w_n)|=\sqrt{\det(A)^2}$, which is easily found from an earlier result from the book.

  • $\begingroup$ Two comments: I don't know where you're getting $Av\cdot A^\top v$. But you're right that he has an error in his formula. Just expand with bilinearity of the inner product. For the second part, think (or read up) about the Gram determinant. $\endgroup$ May 27, 2023 at 18:14
  • $\begingroup$ Certainly $A$ needn't be symmetric, but $\det(A^\top) = \det(A)$, right? $\endgroup$ May 27, 2023 at 18:32
  • $\begingroup$ I’ve actually written up an answer Existence and Uniqueness of a Volume Element on an Oriented, Pseudo Inner Product Space? I suggest focusing on proof #2, because that’s the more naive one, and closer to what Spivak had in mind. Also, if you prefer, ignore all instances of “pseudo” in my question and answer. Modulo some minor notational differences, you should be able to piece together what’s going on. $\endgroup$
    – peek-a-boo
    May 27, 2023 at 19:08
  • $\begingroup$ @TedShifrin 1. Yeah in hindsight my notation wasn't super clear. Basically what I'm trying to say is that if some vector is defined as $Av$, then when taking the transpose with itself the result in terms of $v$ is $v^TA^TAv$, but since Spivak uses $A^TA=I$ earlier I tried to use a halfway point which ended up being more confusing. $\endgroup$ May 27, 2023 at 23:28
  • $\begingroup$ @TedShifrin Yes but if you need to get the value of $T(w_i,w_j)$ given by Spivak it does. $\endgroup$ May 27, 2023 at 23:29


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