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I'm currently working through J. Lee's "Intro to Smooth Manifolds", this is problem 7-22c. It asks the reader to show that $\langle p,q\rangle=\frac{1}{2}(p^*q+q^*p)$ defines an inner product on $\mathbb{H}$. It is trivial to verify that this is linear in the first argument, and satifies conjugate symmetry. However, I think I have misunderstood something. Because $\mathbb{H}$ is a 4-dimensional algebra over $\mathbb{R}$, as stated in the text, so I thought the inner product would have to take values in $\mathbb{R}$.

When verifying the non-degeneracy, I realised that, if $p=(a,b)$ with $a,b\in\mathbb{C}$ (this is the way quaternions are defined in the text), we get $\langle p,p\rangle=(|a|^2+|b|^2,0)$.
Since this does not technically take its values in $\mathbb{R}$, is this really an inner product? It is clearly true that $\langle p,p\rangle=0$ iff $p=0$, and $|a|^2+|b|^2\geq 0$, so we could just consider the first entry of this resulting value of $\langle p,p\rangle$, as the second is always zero, but this doesn't seem quite right to me. What am I missing here?

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There is a copy of $\mathbb{R}$ sitting inside $\mathbb{H}$ consisting (when you take the 4-dimensional $\mathbb{R}$-algebra view) of the scalar multiples of $1$.

Now this borders on the philosophical, but one can think about this copy of the real numbers as the 'actual' real numbers and the problem disappears.

It is like when you believe that natural numbers, integers, rational numbers, real numbers and complex numbers 'exist' in some real-world-sense, it makes perfect sense to say that 'some rational numbers are also integers' or 'all reals are complex numbers but not the other way around' etc.

By means of analogy: you can define the rational numbers formally as equivalence classes of pairs of integers, where the equivalence relation is $(a, b) \sim (c, d)$ iff $ad = bc$. Now this is not very insightful but it works. Now if you do that and I think about rational number in the grade school way then we won't have any problems communicating about and doing computations in the rationals together.

But now suppose that, surprisingly, the outcome of our computation is 2 and I say 'hey that is surprising: this is not just any rational number but actually an integer' then you could, if you felt that way, argue: 'what the hell are you talking about?! An integer is just one element of $\mathbb{Z}$ and $2$ is an infinite subset of the set of pairs of such elements! They are completely different!'

You would be right in a sense, but still I feel I would not be wrong. The situation is the same here where the book thinks about the real multiples of 1 inside $\mathbb{H}$ (so formally the elements of the form $(a, 0)$ with $a$ real) as the 'actual' real numbers we all know and love.

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  • $\begingroup$ Alright, fair enough then. So once I have to define the norm induced by this inner product, I will only consider the first entry and not the second. That somehow feels like "cheating", but I guess that is the only way to do it, thanks. $\endgroup$ Oct 3, 2020 at 13:20
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    $\begingroup$ This answer is exactly right. I'll just point out that in that same problem, I included the remark "Real quaternions can be identified with real numbers via the correspondence $x \leftrightarrow x \mathbb 1$." $\endgroup$
    – Jack Lee
    Oct 3, 2020 at 15:38
  • $\begingroup$ Wait, you are the author of the book? That is really funny. Sometimes I forget that the internet is part of the real world, but then there are nice comments like this that remind me... $\endgroup$
    – Vincent
    Oct 3, 2020 at 21:15

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