On p.154 in Husemoller's Fibre Bundles, during his introduction of Clifford algebras, I found a claim which seems questionable to me (highlighted in red):
You can click here for some context (e.g. the definition of "orthogonal multiplication") but assuming I've expanded everything correctly, here's what I think the remark is saying:
A module $M$ over this Clifford algebra is the data of a real vector space $M$ with a choice of $u_1,\ldots,u_{k-1}\in\mathrm{GL}(M)$ satisfying $u_i^2=-1$ and $u_iu_j+u_ju_i=0$ for $i\neq j$.
Given such an $M$, there is an inner product $(x|y)$ on $M$ and an isomorphism of inner product spaces $\varphi:M\to\mathbb{R}^n$, where $\mathbb{R}^n$ has the Euclidean inner product, such that the $\varphi\circ u_i\circ \varphi^{-1}$ are elements of $\mathrm{O}(n)$.
My confusion is that while the inner product $(x|y)$ on $M$ is certainly constructed so as to ensure that the linear transformations $u_1,\ldots,u_{k-1}$ are orthogonal for that inner product, I see no reason for that inner product to be equivalent to the/an Euclidean inner product on $M$.
In other words, if the inner product $(x|y)$ has signature (Wikipedia link) equal to $(p,q)$, then I agree we can find an isomorphism $\varphi:M\to\mathbb{R}^n$ such that the $u_i$ get sent to elements of $\mathrm{O}(p,q)$ (Wikipedia link), but I see no reason we can ensure $(x|y)$ has signature $(n,0)$.
Now, the definition of $(x|y)$ depends on an original choice of inner product $\langle x|y\rangle$ on $M$, so I suppose it's possible we can always find some $\langle x|y\rangle$ such that $(x|y)$ has the right signature.
If that's the case, can someone provide a proof? Or is the remark incorrect? Or have I misinterpreted the remark?
If any of the terms are unclear, I'm happy to provide more screenshots from the book.
Thanks in advance for any help.
I think I was just being overly cautious here.
The starting inner product $\langle x|y\rangle$ on $M$ is positive definite, i.e. of signature $(n,0)$, simply by the definition of "inner product".
For each $\sigma\in \Gamma_{kl}$, the form $\langle \sigma(x)|\sigma(y)\rangle$ on $M$ is positive definite, again by simple observation: for any $x\in M$ $\sigma(x)$ is again an element of $M$, so $\langle \sigma(x)|\sigma(x)\rangle>0$. This is I think what was the sticking point for me before - for whatever reason I was convinced there was no guarantee $\langle \sigma(x)|\sigma(y)\rangle$ would necessarily be positive definite.
A sum of positive definite matrices is positive definite, so since $\langle \sigma(x)|\sigma(y)\rangle$ is positive definite (and symmetric) for any $\sigma\in\Gamma_{kl}$, their sum $(x|y)$ is as well.
All inner products on a finite-dimensional $\mathbb{R}$-vector space are equivalent, via the obvious argument (take an orthonormal basis for one inner product and define a map by sending it to an orthonormal basis for another inner product)