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I just started studying Clifford algebras and I am puzzled by the following example.

Let $X$ be a Hilbert space with $\mathrm{dim}\ X = 1$. Let $\{e_1\}$ be the basis for $X$. Then the Clifford algebra $\mathcal{C}(X)$ consists of all elements of the form $\alpha + \beta e_1$ where $\alpha,\beta \in \mathbb{R}$. Hence, $\mathrm{dim}\ \mathcal{C}(X) = 2$.

However, when I try taking $X = \mathbb{R}$ then $\mathrm{dim}\ X = 1$ and all elements of $\mathcal{C}(X)$ are of the form $\alpha + \beta e_1 \in \mathbb{R}$. But then $\mathrm{dim}\ \mathcal{C}(X) = 1$ ?

Konrad Burnik
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    When you use a basis of the Hilbert space to compute the associated Clifford algebra, you are writing the latter as an algebra generated by the basis vectors with some constraints added. This basically means that $e_1$ is a formal element. That it used to be $1 \in\mathbb R = X$ does not mean that $1 + e_1 = 2$ in $\mathcal C(X)$. – SolubleFish May 01 '21 at 11:09
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    Yes, when you write $\alpha$ you mean $\alpha \cdot 1$ with $1$ the identity of the Clifford algebra, while $e_1$ is an element such that $e_1^2 = -1$. I studied Clifford algebras on "Representations of compact Lie groups" of Bröcker and Dieck, maybe this reference can be helpful to you. – Lorenzo Panebianco May 01 '21 at 11:17
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    The product $\beta e_1$ should rather be seen as a tensor product $\beta\otimes e_1$. This means that they factors are considered to belong to different linear spaces, but any factor can be moved from one to the other: $(c\beta) \otimes e_1 = \beta \otimes (c e_1)$. – md2perpe May 01 '21 at 13:04

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This is a bit tricky to answer without knowing your background, so I'll try to err on the side of a more accessible answer.

Basically, "$\mathbb R$ as a vector space" is a different thing from $\mathbb R$ (where numbers can be multiplied), and if you keep them separate you won't have this kind of issue. One choice, but not the only one, would be to distinguish them in notation by writing "$\mathbb R$ as a vector space" as "$\mathbb R^1$" (as discussed in Is $\mathbb{R}^1$ really the same as $\mathbb{R}$?). Let's say that when we have elements of "$\mathbb R$ as a vector space" (or $\mathbb{R}^1$), we write them in angle brackets so that $-2\langle-1\rangle=\langle2\rangle$ and $\langle-1\rangle+\langle2\rangle=\langle1\rangle$, but the product $\langle3\rangle\langle4\rangle$ is not defined since you can't just "multiply" vectors. There is also a natural inner product where $\langle2\rangle\cdot\langle3\rangle=6\ne\langle6\rangle$. With this ($\mathbb{R}^1$) as our Hilbert space $X$, elements of $\mathcal{C}(X)$ would be like $2+3\langle1\rangle\notin\mathbb R$.

It may be easier to understand the the $\mathcal{C}(X)$ example with $X$ being a one-dimensional vector space that is more clearly different from $\mathbb R$, like let $X$ be "the vector space of horizontal arrows, with addition given by attaching arrows tail-to-tip, and scalar multiplication given in the standard way where multiplication by $2$ doubles the length, multiplication by $-1$ flips the direction, etc." We can define $|\mathbf{v}|$ to be the length of $\mathbf v$. Then $X$ is a Hilbert space where $\mathbf v\cdot\mathbf w$ is given by $\mathbf v\cdot\mathbf w=\pm|\mathbf{v}||\mathbf{w}|$ with the minus sign when $\mathbf v$ and $\mathbf w$ point in opposite directions.

Any arrow of nonzero length could be a basis vector for $X$, say $e_1=``\to"$ (so that $-e_1=``\leftarrow"$, etc.). Then $\mathcal{C}(X)$ consists of all expressions of the form $\alpha +\beta e_1$, like $``3+2\to"=``3+\longrightarrow"$.

Mark S.
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  • Thanks for your answer! Can we just look at $\alpha + \beta e_1$ as an ordered pair $(\alpha, \beta)$ w.r.t. basis $(1,e_1)$? Similar how we look at complex numbers $\alpha + \beta \mathbb{i}$ as ordered pair $(\alpha, \beta)$ w.r.t. basis $(1, \mathbb{i})$? – Konrad Burnik May 02 '21 at 16:10
  • @Math Curious 001, yes it can, but doing it that way depends on the choice of basis vector $e_1$ in a way that the standard way of writing things in the Clifford Algebra does not. – Mark S. May 02 '21 at 16:14
  • So, when looking at $\mathcal{C}(\mathbb{R})$ just as a bare set, what exactly is that? – Konrad Burnik May 02 '21 at 20:03
  • @MathCurious Depends on several set theory choices. If could certainly be the set of ordered pairs of reals. But as I mentioned in my answer, failing to distinguish the reals (what $\alpha$ and $\beta$ can be) from your one-dimensional vector space risks confusion. If $X$ is my one-dimensional vector space of arrows, then you could take $\mathcal{C}(X)$ to be $\left{(\alpha,\mathbf v):\alpha\in\mathbb R,\mathbf v\in X\right}$. Only if you choose a basis vector $e_1$ could you replace this with the set of pairs of reals. – Mark S. May 02 '21 at 21:34
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    I think the problem was in taking $X = \mathbb{R}$ in the first place. I think the right thing to do here is to take $X = \mathbb{R}^1$ instead since there is a subtle difference between $\mathbb{R}^1$ and $\mathbb{R}$ as described here: https://math.stackexchange.com/questions/4124752/is-mathbbr1-really-the-same-as-mathbbr

    According to this $\mathbb{R}$ as a vector space over $\mathbb{R}$ is in fact $\mathbb{R}^1$ and that should clear up this confusion then.

     – Konrad Burnik May 03 '21 at 17:58
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    @MathCurious Yeah, that's what I was getting at with the "angle brackets" stuff at the beginning. – Mark S. May 03 '21 at 18:12
  • Ok, I think if you just add a reference to https://math.stackexchange.com/questions/4124752/is-mathbbr1-really-the-same-as-mathbbr to the answer I can accept it. – Konrad Burnik May 03 '21 at 18:16