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I am physicist studying the Dirac algebra as a formal algebraic structure in abstract algebra. I am not well-trained in modern algebra, but I am self-studying. This post is long-ish because I will write what I have (hopefully) understood, and then I will ask a couple of questions about the remaining gaps in my understanding. Any corrections and/or clarifications will be much appreciated. The main gist is that the Dirac variant of Clifford algebra has 16 elements, and every Clifford algebra has $2^d$ elements where $d$ is the dimensionality of $V$. The Dirac algebra is the anti-commutator relation

$$ \{\gamma^\mu,\gamma^\nu\}=2\eta^{\mu\nu}\boldsymbol1, $$

but this looks like the generating vector space $V$ must have $d=5$ since $\gamma^\mu$ and $\boldsymbol1$ are five linearly independent matrices. However, the Clifford algebra in question does not have $2^5$ elements. Now I will layout my thinking.

A Clifford algebra $\mathcal{C}\!\ell(V,Q)$ is generated by a vector space $V$ equipped with a quadratic form $Q$. The dimension of the Clifford algebra is $2^d$ where $d$ is the dimension of $V$. In general, an algebraic structure is a set $S$ equipped with a bilinear operation on every element of $S\times S$. An algebra is a sort of algebraic structure in which the set $S$ is equipped as a vector space $V$, and the binary operation on $S\times S$ must have $S$ itself as its target space. Substituting $V$ for $S$, an algebra is a vector space $V$ equipped with a binary operation $\circ$ such that

$$ \circ:V\times V\to V. $$

The scalar product on a vector space (making it a inner or scalar product space) does not support the definition of an algebra because

$$ \langle\,,\rangle:V\times V\to K. $$

where $V$ is implicitly a vector space over the field $K$. Now we come to the part where things cease to be crystal clear. I think the Clifford algebra is also a $2^d$-dimensional vector space. I think this is obvious, but I am not seeing any authors make this point. Am I wrong? If I am right but it is not mentioned, why don't we use this language? Let us call the space spanned by the elements of $\mathcal{C}\!\ell(V,Q)$ by the name $V_{\mathcal{C}\ell}$. Now, for context, I remind that I am looking at the Dirac algebra defined by

$$ \{\gamma^\mu,\gamma^\nu\}=2\eta^{\mu\nu}\boldsymbol1, $$

where $\boldsymbol1$ is the multiplicative identity. Let us call the 4-dimensional vector space of $\gamma$-matrices by the name $V_\gamma$. The anti-commutator bracket satisfies the $S\times S\to S$ algebra condition, despite $\boldsymbol1$ not being a linear combination of $\gamma^\mu$, because the case of $\{\gamma^\mu,\boldsymbol1\}$ is implicit from $(\gamma^\mu)^2=\boldsymbol1$ as

\begin{align} \{\gamma^\mu,\boldsymbol1\}&=\{\gamma^\mu,(\gamma^\mu)^2\}\\ &=(\gamma^\mu)^3+(\gamma^\mu)^3 \\ &=\big[(\gamma^\mu)^2+(\gamma^\mu)^2\big]\gamma^\mu\\ &=\{\gamma^\mu,\gamma^\mu\}\gamma^\mu\\ &=2\eta^{\mu\mu}\boldsymbol1\gamma^\mu~~. \end{align}

Although the algebra has only defined the anti-commutator, the identity element is defined to have the property $\boldsymbol1 \mathsf{M}= \mathsf{M}\boldsymbol1 =\mathsf{M}$, so I can say that the operation on the last line is defined. Thus, the Dirac algebra is defined with the anti-commutator on the 5D vector space $V\oplus K\boldsymbol1$ where $K\boldsymbol1$ is a 1-dimensional vector space:

$$ \{\,,\}:(V\oplus K\boldsymbol1)\times(V\oplus K\boldsymbol1)\to V\oplus K\boldsymbol1 $$

Now, really getting to my question, this binary operation is defined on neither $V_\gamma\times V_\gamma$ nor $V_{\mathcal{C}\ell}\times V_{\mathcal{C}\ell}$ (where $V_{\mathcal{C}\ell}$ is the 16-dimensional space spanned by the linearly independent elements of $\mathcal{C}\!\ell(V,Q)$).

$\bullet~$ A lot of material seems to suggest that the Dirac algebra is generated from the 4-dimensional space of $\gamma$-matrices, but that conflicts with the requirement of an algebra to have $\circ:V\times V\to V$ since $\boldsymbol1$ is not a linear combination of $\gamma$-matrices. If the dimension of the Dirac algebra is $2^d=16$, then $V$ must be 4-dimensional, but that makes $\{\gamma^\mu,\gamma^\nu\}=2\eta^{\mu\nu}\boldsymbol 1$ not in the form $\circ:S\times S\to S$. What is the error in my thinking when I feel like $V$ must be 5-dimensional?

$\bullet~$ What language is used to differentiate the 16-dimensional vector space from the 4- or 5-dimensional one on which the anti-commutator is defined?

$\bullet~$ Why am I not seeing anyone say clearly in their expositions of Clifford algebras that the dimension of the Clifford algebra is the dimension of the vector space spanned by the algebra's elements? Is the algebra, in fact, a vector space? The elements appear to satisfy the vector space axioms, as far as I can tell, but no (few?) authors use this language.

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    $\begingroup$ " Is the algebra, in fact, a vector space" Any algebra is a vector space, I think that is a point which is considered elementary enough that no author would feel the need to point it out in the case of Clifford algebras. $\endgroup$ Commented Apr 10 at 13:33
  • $\begingroup$ @rschwieb The clifford algebra is "generated by a vector space $V$" A vector space does not have multiplication of vectors defined, so I cannot understand your answer. If you were the one who closed my question as a duplicate, please re-open it because this is a completely different question than the one you helped me with earlier. This question has literally nothing to do with the other question other than the common theme of the Clifford algebra. Thank you. $\endgroup$
    – Nada Band
    Commented Apr 10 at 13:36
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    $\begingroup$ To be clear, I think bullet points 1 and 2 are explicitly addressed in the other question's solutions. The third one is not, but the answer is mundane and something we've handled here. $\endgroup$
    – rschwieb
    Commented Apr 10 at 13:41
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    $\begingroup$ @NadaBand OK, I will just take a mulligan this time in the spirit of flexibility. I'll retract and try to answer your question even though it seems very similar to me. $\endgroup$
    – rschwieb
    Commented Apr 10 at 13:46
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    $\begingroup$ @NadaBand Also, it is fine if, faced with a duplication closure, to add content to explain explicitly how the two things are different. But what you added was simply a long paragraph saying over and over again "this is not the same. It is totally different. It is self-evidently different. I can't believe this is happening." The lack of substance will probably annoy readers. If there really were a difference you should take the time to explain it. $\endgroup$
    – rschwieb
    Commented Apr 10 at 14:21

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I think the Clifford algebra is also a $2^𝑑$-dimensional vector space. I think this is obvious, but I am not seeing any authors make this point. Am I wrong?

This is like the first thing you learn about Clifford algebras and seems to be abundantly said in expositions I have read. A good litmus test is that wikipedia includes the fact.

A lot of material seems to suggest that the Dirac algebra is generated from the 4-dimensional space of 𝛾 -matrices

It is, as an algebra, generated by those four matrices, yes. (Points 1 and 2 here)

but that conflicts with the requirement of an algebra to have ∘:𝑉×𝑉→𝑉 since 1 is not a linear combination of 𝛾-matrices.

No it doesn't. 1 is indeed not a linear combination of $\gamma$s (nobody said it was.) It is simply not in the span (as a vector space) of the $\gamma$s. But it is generated by the algebra operations, because this is a linear combination of products of elements of the algebra: $\frac12 \{\gamma^i,\gamma^j \}=1$ for some particular $i,j$ (I'm not handy enough to have a pair that works in mind.)

You can take a subset of a vector space and find its subspace which is what it generates as a vector space. You can take a subset of an $F$ algebra and find its subalgebra which is what it generates as an $F$ algebra. In this latter case you can also take the same subset and ask about its span too, because it is a fortiori a vector space too. But the subspace and subalgebra of course can be different.

What language is used to differentiate the 16-dimensional vector space from the 4- or 5-dimensional one on which the anti-commutator is defined?

Again, this is all about whether you're generating the object as a vector space or as an algebra, same as in the linked solution above. The algebra is a 16 dimensional vector space too, generated by the elements of a four dimensional subspace that we initially started with. You can also ask about the 5 dimensional subspace generated by $V$ with $1$ inside of the algebra but AFAICT it is of no importance.

Why am I not seeing anyone say clearly in their expositions of Clifford algebras that the dimension of the Clifford algebra is the dimension of the vector space spanned by the algebra's elements?

IMO the dimension of an $F$-algebra (Clifford or not) is defined in every textbook this way. If it is not in surveys you have been reading it may well be because the authors overlooked the point as trivial.

can you please explain how the anti-commutator is defined as 𝑉×𝑉→𝑉 with 𝑉 4-dimensional so the Clifford algebra has dimension $2^𝑑=16$ , or cite the error in my thinking that the algebra must be defined on every element of the Cartesian product of 4-dimensional vector space with itself, mapping to the 4-dimensional target space 𝑉 ?

Quoting my other answer and adding emphasis:

It turns out to be 16 dimensional not because of $4^2=16$, but because all Clifford algebras constructed this way have dimension a power of $2$, and in this case it's $2^4=16$. It's not because you're doing pairwise products between the $\gamma$'s, it's because you can take any of the $2^4$ subsets of those four matrices, and form a product with increasing indicies, and that gives you 16 things, and they happen to be a basis (as a vector space) for the algebra. Any other finite product of these matrices can be rewritten as a linear combination of those 16 matrices after using the relations, so they are generated by those 16.

Maybe perhaps an issue for you is thinking of the commutator bracket as being an operation. That is not the case here. While $\{\cdot,\cdot\}$ is being used as a shorthand here for an actual multiplication and addition/subtraction in the algebra, it is not an operation defined $V\times V\to V$, it could only be considered a bilinear map $V\times V\to C\ell(V)$ (not any sort of "operation.")

Anyhow, we don't really care to think of it as an operation. All we care about is that the collection $X=\{(\{\gamma^\mu,\gamma^\nu\}-2\eta^{\mu\nu}\cdot\mathbf{1})\mid\mu,\nu\}$ is a subset of the free algebra generated by $V$, and when we take the quotient algebra by the ideal generated by $X$, it creates this useful algebra $C\ell(V)$.

Finally, if an algebra is a vector space 𝑉 equipped with a binary operation 𝑉×𝑉→𝑉 , can you please clarify, finally, what 4-dimensional 𝑉 satisfies this requirement for 𝑉×𝑉→𝑉 ? It can't be the space of 𝛾 -matrices since that is not the target space. I think this must be terribly stupid on my part for me to still not be getting it, but I cannot see what 𝑉 satisfies this

If $V$ denotes any vector space, the algebra $C\ell(V)$ is a 16 dimensional vector space with an operation $C\ell(V)\times C\ell(V)\to C\ell(V)$. Could the issue be as simple as you confusing $V$ with $C\ell(V)$ somehow?

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  • $\begingroup$ Thank you for taking the time. I was just looking at the Wiki you linked, and that section only mentions the dimension of the algebra, not that it was a vector space. The absence of such an explicit statement left me wondering why they might leave it unstated. Still, I am glad my thinking was correct, so thank you. $\endgroup$
    – Nada Band
    Commented Apr 10 at 14:19
  • $\begingroup$ Finally, if an algebra is a vector space $V$ equipped with a binary operation $V\times V\to V$, can you please clarify, finally, what 4-dimensional $V$ satisfies this requirement for $V\times V\to V$? It can't be the space of $\gamma$-matrices since that is not the target space. I think this must be terribly stupid on my part for me to still not be getting it, but I cannot see what $V$ satisfies this. $\endgroup$
    – Nada Band
    Commented Apr 10 at 14:19
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    $\begingroup$ @NadaBand Ok well, i added an answer to that too. $\endgroup$
    – rschwieb
    Commented Apr 10 at 14:31
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    $\begingroup$ Yes, the issue was that I was confusing $V$ with $\mathcal{C}\!\ell(V)$. The binary operation (or map) is defined on the elements of the Clifford algebra, not on the elements of the vector space that generate Clifford algebra (which is 4-dimensional while the Clifford algebra is a 16-dimensional vector space.) Now I get it. Thanks for taking the time! $\endgroup$
    – Nada Band
    Commented Apr 10 at 14:45

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