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I apologize in advance for a rather wordy question.

As a physicist trying to learn new mathematics, I figured this was the place to ask. I am having trouble understanding the algebraic formulation of Clifford algebras and I am (hopefully) looking for some clarity.

I will say from the get-go, that my understanding of quotient spaces and ideals are elementary. By this I mean, that I understand the basic definitions at face-value and some of the simpler examples, though both subjects has always been somewhat confusing to me. So forgive my ignorance in advance.

What I understand:

Up until now, my understanding of Clifford algebras is the following (I'll be brief and omit details):

If $(V,g)$ is a vector space, equipped with a symmetric, bilinear form $g : V \times V \rightarrow \mathbb{F}$, then the Clifford algebra $\text{C}\ell (V,g)$ of $V$ is the algebra spanned by the elements of $V$, subject to the condition that $v^2=g(v,v)$.

From this, I can deduce and understand that $\text{C}\ell^{\pm} (\mathbb{R}^n)$ is the algebra generated by the vectors subject to $v_i v_j +v_j v_i = \pm 2 \delta_{ij}$. I understand also that the Clifford product of any two vectors $u,v$ may be written as $uv=u \cdot v + u \wedge v$, where $u \cdot v$ is the inner product and $u \wedge v$ is the wedge product. From this, the representations of the various Clifford algebras follows (both in the real and complex cases) and I'm pretty happy.

What I don't understand:

Now, if I move on to more advanced texts, I often come across the following:

If $T(V)$ is the tensor algebra $T(V) := \bigoplus_{k=0}^{\infty} V^{k \otimes}$, with $T^0(V) = \mathbb{F}$, then the Clifford algebra $\text{C}\ell(V,g)$ is the quotient $\text{C}\ell (V,g)=T(V)/I(V)$ of $T(V)$ by the ideal $I(V)$, generated by the elements of the form $v \otimes u + u \otimes v -2g(u,v)$.

Though I know what all the ingredients are, I simply do not see the correlation between this definition and the more "geometric" notion of a Clifford algebra! All my intuitions about the subject are gone.

From my (again, very basic) understanding of quotient spaces, $\text{C}\ell (V,g)$ should be the space of equivalence classes, where $u \sim v$ iff $u = v+w$ for some $w \in I(V)$.

Working from the definition of the ideal, I can deduce that $u \otimes v \sim \frac{1}{2}(u\otimes v-v \otimes u)+g(u,v) = u \wedge v + g(u,v)$, but I cannot get much further and it does not do any wonders for my understanding.

I will try to be concrete:

Questions:

1) How is the "algebraic" definition of the Clifford algebra $\text{C}\ell(V,g)$ (above) related to the concept of the Clifford algebra as the algebra generated by elements in $V$, subject to the condition that $v^2=g(v,v)$ (in the most general sense)?

2) How does one obtain the Clifford product, now that the Clifford algebra is a space of equivalence classes?

3) Why is this definition of the Clifford algebra "useful", as opposed to the geometric formulation?

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  • 2
    $\begingroup$ Question : how would you define "the algebra spanned by $V$ and subject to the condition that $v^2=g(v,v)$" ? I mean, $V$ is not a subset of an algebra, so the algebra spanned by $V$ should have a precise meaning, as well as the "condition that ...". Note also that if such an algebra exists, it is in general not unique. The answer is simply $T(V)/I(V)$. Indeed, $T(V)$ is the free algebra spanned by $V$. $I(V)$ is the conditions we want to impose on our algebra. Now, in $T(V)/I(V)$, one has $uv+vu-g(u,v)=0$ since this is an element in the ideal. Hence $uv+vu=g(u,v)$. $\endgroup$ – Roland Dec 30 '18 at 23:26
  • $\begingroup$ Thank you! See also my comments on user3482749's response :) $\endgroup$ – Henrymerrild Dec 31 '18 at 15:31
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I'm also going to start with a bit of a disclaimer: my knowledge of Clifford Algebras is fairly elementary (I know the algebraic definition, I have a vague idea that they're useful in physics in some way and that quantum group theorists all care about them, and they're occasionally interesting examples of things that I do care about), but I do know a fair amount about tensor products, tensor algebras, quotient spaces and the like, so hopefully we can get somewhere between us.

  1. I think it might be most useful here if I start with an example, I'll use the same one that you used: $\mathcal{Cl}^{\pm}(\mathbb{R}^n)$. To show that the two conditions being imposed are equivalent, use the Polarization Identity: $u\otimes v + v\otimes u = 2g(u,v) = g(u+v,u+v)-g(u,u)-g(v,v)$, so taking $u = v$ gives $v\otimes v = \frac{1}{2}(g(2v,2v)-2g(v,v)) = g(v,v)$ (the other direction is by a proof identical to the one that you used for the $\mathcal{Cl}^\pm(\mathbb{R}^n)$ example). I think the key thing to notice here is that your "subject to the condition that..." is already a quotient: quotienting out by the ideal generated by a set $S$ is exactly the same as imposing the condition that all elements in $S$ are zero. In this case, your $v^2 = g(v,v)$ condition corresponds exactly to quotienting out by the ideal generated by all elements of the form $v^2 - g(v,v)$, which that proof mentioned above will show is precisely the ideal in your algebraic definition.

  2. Essentially, exactly the same as in the other case: we have a distinguished copy of $V$ living inside $\mathcal{Cl}(V,V)$, and we can extend $g$ from that to the whole of $\mathcal{Cl}(V,V)$ (this is yet another thing that breaks in the characteristic 2 case).

  3. Basically, because we know an awful lot about tensor products and quotient algebras, and it's kinda useful to be able to make use of that. Also, because it is, in many ways, more concrete than your geometric definition: whereas you say "the [free-est] algebra generated by the elements of $V$", this definition simply substitutes in what that free-est algebra is: it's the quotient of the free algebra generated by $V$ by the ideal associated to that condition, and the free algebra generated by $V$ is exactly that infinite direct sum of tensor products of $V$ in the algebraic definition. So all that this definition is doing is taking your definition, and writing down what all of the things in it actually are.

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  • $\begingroup$ Thank you for your thorough answer! It already made the whole thing more tangible. I only have a few follow up comments/questions: 1) The crux seems to be that “quotienting out by the ideal generated by a set S is exactly the same as imposing the condition that all elements in S are zero.” This was originally the source for my confusion (probably due to the fact, that my grasp on quotient spaces is very limited). If I understand this correctly, the ideal is “squashed” to 0 and this is then extended to the rest of the space? [continued] $\endgroup$ – Henrymerrild Dec 31 '18 at 15:29
  • $\begingroup$ In that case, I see that the ideal has exactly the form (albeit with tensor products) that we want to impose on our algebra. 2) I am not sure I follow this exactly. In the definition of the ideal, we are using tensor products $\otimes$; but I assume that we want an identification $uv+vu-2g(u,v)=0$, where the multiplication is the Clifford product (as a product on the quotient space)? $\endgroup$ – Henrymerrild Dec 31 '18 at 15:29
  • $\begingroup$ 3) I see! Once again, thanks for the effort (and happy new years)! $\endgroup$ – Henrymerrild Dec 31 '18 at 15:29

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