Aren't these linear transformations only orthogonal with respect to *some* inner product?

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 see no reason for that inner product to be equivalent to the/an Euclidean inner product on M.

It seems as if you are interpreting "orthogonal" to be "Euclidean," which is probably the problem.

The modifier "orthogonal" is used to describe any geometry which preserves a symmetric bilinear form.

So what I think is going on is that they are claiming there exists a bilinear form for which the $u_i$ all lie in the symmetry group for the form.

• Thanks for your answer - however, I think I was ultimately worrying over nothing. – Sal Jan 19 '15 at 3:59

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)