$\newcommand{\Cl}{\mathscr{Cl}(V)}$$\newcommand{\ext}{\Lambda(V)}$Let $V$ be a finite dimensional vector space over a field with characteristic not equal to two. Assume we have made a choice of symmetric bilinear form for $V$.
It is known that the Clifford algebra $\Cl$ and the exterior algebra $\ext$ are isomorphic as vector spaces (this is an exercise in Greub's Multilinear Algebra, I believe it is mentioned in Wikipedia, and also see this and this and this related question on Math.SE).
Now obviously the Clifford algebra $\Cl$ with the Clifford product is not isomorphic as an algebra to the exterior algebra $\ext$ with the wedge product.
However, the wedge product is also defined on $\Cl$ and different from the Clifford product.
Question: Is the Clifford algebra $\Cl$ with the wedge product, i.e. not with the Clifford product, isomorphic as an algebra to the exterior algebra $\ext$ with the wedge product?
Attempt: I feel like this should follow immediately from the following two facts:
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The Clifford algebra $\Cl$ and the exterior algebra $\ext$ are isomorphic as vector spaces.
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A list of vectors $(v_1, \dots, v_n)$ is linearly independent if and only if its wedge product is non-zero: $$v_1 \wedge \dots \wedge v_n \not=0\,.$$
(I believe this second fact is correct, it is the content of at least one exercise in Lee's Introduction to Smooth Manifolds if I remember correctly, although I haven't gotten around to doing it yet.)
Note: This is effectively a follow-up to this question I asked on MathOverflow.
Wedge product on the Clifford Algebra This might not be correct; if so, please explain why and/or give a pointer to a reference which explains why and I will also accept your answer.
(Note: what I had previously was definitely wrong, so here's a second attempt, again based on this document.) We can decompose the Clifford product $vw$ as follows: $$vw = \frac{1}{2}(vw + wv) + \frac{1}{2}(vw -wv) \,. $$ Noting that the first term on the RHS is symmetric and bilinear (in $v$ and $w$), the author Chisholm states that it is plausible that this could be the inner product. (See the top of p.4) I think if one takes the formal definition of Clifford algebra, it follows from the quotient relation $v^2 = \langle v, v \rangle$ that this makes sense. So then the second term on the RHS, $\frac{1}{2}(vw -wv)$, is denoted $v \wedge w$, one has from the definition that it is skew-symmetric. In the Euclidean case for $\mathbb{R}^3$, it follows from $\frac{1}{2}(vw+wv)=\langle v,w\rangle=|u||v|\cos\theta$ that $(v \wedge w)^2 = -|v|^2|w|^2 \sin^2\theta = -|v \times w|^2$.
Also, the wedge product of any element of the Clifford algebra $\Cl$ with a scalar is just scalar multiplication (I think).
I think the definition of the wedge product above generalizes so that for any $v_1, \dots, v_n \in V$: $$v_1 \wedge \dots \wedge v_n = \frac{1}{n!}\sum_{\sigma \in S_n} (\operatorname{sgn} \sigma) v_{\sigma(1)} \dots v_{\sigma(n)} \,, $$ where the multiplication is the Clifford product (see eq. (30) p.13 here) but I'm not sure.
If that formula is correct, then the answer to my question might automatically be affirmative due to a bijection between the Clifford algebra (with the wedge product) and alternating/skew-symmetric tensors of rank $\le n$. In other words the proposed formula is very similar to that given for creating the exterior algebra from the tensor algebra.
Following Qiaochu Yuan's suggestion in the comments, the wedge product of three vectors would for example then be: $$u \wedge v \wedge w = \frac{1}{6}uvw + \frac{1}{6}vwu + \frac{1}{6}wuv -\frac{1}{6}vuw -\frac{1}{6}uwv -\frac{1}{6}wvu \\ = \frac{1}{6}u(vw – wv)+\frac{1}{6}v(wu-uw)+\frac{1}{6}w(uv-vu) \\ = \frac{1}{3}u(v \wedge w) + \frac{1}{3}v(w \wedge u) + \frac{1}{3}w(u \wedge v) \,.$$
My hope is that this would be enough to define (via induction) what is meant by the wedge product of arbitrary elements of the Clifford algebra $\Cl$.
Basically the motivation for my claim/belief that the Clifford algebra $\Cl$ also has a wedge product is how many texts on the Clifford algebra of $\mathbb{R}^n$ with the Euclidean inner product treat the Clifford product (for vectors) essentially as a combination of the inner and wedge products, i.e. an extension of the wedge product.
This seems like it might be compatible with the formal definition of Clifford algebra, based on the fact that one should have $v^2 = vv = \langle v, v \rangle$, but I was hoping/assuming someone here might have already learned all of this and can confirm/deny that this is true before I try to verify it all in detail by myself only to find out that I have been trying to prove something false.
Wikipedia says that:
…if one takes the Clifford algebra to be a filtered algebra, then the associated graded algebra is the exterior algebra.
To be honest I don't know what that means, but it seems like it might be stronger than: "the Clifford algebra is isomorphic to the exterior algebra as a vector space".
Best Answer
$\newcommand{\Cl}{\mathscr{Cl}(V)}$The problem with your definition of the wedge product in $\Cl$ is that it is an $n$-ary product of vectors. And so it's not clear to me how one would even discuss associativity of it (i.e. it doesn't allow one to distinguish between the product of a bivector and a vector vs the product of a vector and a bivector). So while $\Cl$ with the wedge product is an algebra -- it isn't an associative algebra like the exterior algebra is. Let me instead recommend a different definition based on grade projection (basically the same definition given in the works of Alan Macdonald):
To define the wedge product, we first define it in the case of single-grade elements. Let $A,B$ be an $r$-vector and $s$-vector, respectively. Then
$$A\wedge B := \langle AB\rangle_{r+s}$$
Now using this special case, we define the wedge product for all multivectors. Let $C,D$ be (potentially mixed grade) multivectors in $\Cl$. Then we define
$$C \wedge D := \sum_{j,k} \langle C\rangle_j \wedge \langle D\rangle_k$$
Proof that for all $A,B,C\in \Cl$, we have $A\wedge (B\wedge C) = (A\wedge B)\wedge C$:
Consider a $j$-vector $A$, $k$-vector $B$, and $l$-vector $C$. Then $$(A\wedge B)\wedge C = \langle AB\rangle_{j+k}\wedge C = \langle \langle AB\rangle_{j+k}C\rangle_{j+k+l} = \langle (AB)C\rangle_{j+k+l}$$ where that last step holds because only the \langle AB\rangle_{j+k} part of $AB$ can contribute to $\langle (AB)C\rangle_{j+k+l}$.$^\dagger$.
Then $$(A\wedge B)\wedge C = \langle (AB)C\rangle_{j+k+l} = \langle A(BC)\rangle_{j+k+l} = A\wedge\langle BC\rangle_{k+l} = A\wedge(B\wedge C)$$
To finish the proof by allowing $A,B,C$ to be any multivectors in $\Cl$, just note that our definition of the wedge product is bilinear.$^\ddagger$$\ \ \ \ \square$
Now we'll show that with this definition, your proposition holds. Take $\wedge_C$ to be the wedge product in $\Cl$ and $\wedge_T$ to be the wedge product in $\Lambda V$ (i.e. the coproduct of the antisymmetric tensor powers of $V$).
Let $\{e_1, \dots, e_n\}$ be an orthogonal basis for $V$ wrt the symmetric bilinear form and construct bases for $\Cl$ and $\Lambda V$ in the obvious way from it. Let $\Gamma: \Cl \to \Lambda V$ be given by $$\Gamma(A_0 + A_1e_1 + A_2e_2 + \cdots + A_{12\cdots n}e_1\wedge_C e_2\wedge_C \cdots \wedge_C e_n) = A_0 + A_1e_1 + A_2e_2 + \cdots + A_{12\cdots n}e_1\wedge_T e_2\wedge_T \cdots \wedge_T e_n$$
By definition this is linear. It's also clearly invertible. To see that it's multiplicative, it suffices (due to linearity of $\Gamma$ and bilinearity of $\wedge_C$) to show that $\Gamma$ is multiplicative on blades. Let $A=A_{i_1\cdots i_p}e_{i_1}\wedge_C\cdots\wedge_C e_{i_p}$ be a $p$-blade and $B=B_{j_1\cdots j_r}e_{j_1}\wedge_C\cdots\wedge_C e_{j_r}$ an $r$-vector. Then $$\begin{align}\Gamma(A\wedge_C B) &= \Gamma(A_{i_1\cdots i_p}e_{i_1}\wedge_C\cdots\wedge_C e_{i_p}\wedge_CB_{j_1\cdots j_r}e_{j_1}\wedge_C\cdots\wedge_C e_{j_r}) \\ &= A_{i_1\cdots i_p}e_{i_1}\wedge_T\cdots\wedge_T e_{i_p}\wedge_T(B_{j_1\cdots j_r}e_{j_1})\wedge_T\cdots\wedge_T e_{j_r} \\ &= (A_{i_1\cdots i_p}e_{i_1}\wedge_T\cdots\wedge_T e_{i_p})\wedge_T(B_{j_1\cdots j_r}e_{j_1}\wedge_T\cdots\wedge_T e_{j_r}) \\ &=\Gamma(A_{i_1\cdots i_p}e_{i_1}\wedge_C\cdots\wedge_C e_{i_p})\wedge_T\Gamma(B_{j_1\cdots j_r}e_{j_1}\wedge_C\cdots\wedge_C e_{j_r}) \\ &= \Gamma(A)\wedge_T\Gamma(B)\end{align}$$
A linear, invertible, multiplicative map is by definition an isomorphism of algebras. Hence $\Cl$ is isomorphic to $\Lambda V$, as desired.
Another Interesting Tidbit:
$\Lambda V$ is actually (or at least can be) defined by a set of axioms like other types of vector spaces.
From nlab:
It is easily confirmed that $\Cl$ satisfies these properties under the definition of $\wedge$ given above (and thus why associativity is so important for $\wedge$). Thus, not only can we construct an isomorphism, but we can even consider $\Cl$ to just be $\Lambda V$ with extra structure (the Clifford product).
$\dagger:$ To prove this, note that it is known that if $\Bbb F$ is not characteristic $2$, then $V$ has an orthogonal basis wrt any symmetric bilinear form. Then you can construct a "standard" basis for $\Cl$ and decompose your multivectors wrt to it.
$\ddagger:$ By the linearity of the grade projection operator.