Is this exterior algebra?

70 Views Asked by At

I'm learning algebraic topics like exterior algebra and Clifford algebra. I'm trying to understand exterior algebra now or more specifically - different constructions of it. I like the abstract definition of exterior algebra (and other algebras) via universal property. I've seen perhaps 4 different constructions of exterior algebra (including 2 that I, you can say, made up myself) and it made me to start wondering wether I can always define

$$v_1 \wedge \ldots \wedge v_k:=\frac{1}{k!}\sum_{\sigma \in S_k}\mathrm{sgn}(\sigma)v_{\sigma(1)}\cdot \ldots \cdot v_{\sigma(k)}$$

for $v_1,\ldots,v_k$ in some algebra $A$ with some multiplication $\cdot$? By $\cdot$ I mean here some arbitrary algebra multiplication (say: square matrices with matrix multiplication) not necessarily tensor product $\otimes$ of a tensor algebra. Will $A$ with thus defined $\wedge$ multiplication be the exterior algebra of $A$?

1

There are 1 best solutions below

1
On

Not really. Remember that the elements

$$ v_1\land...\land v_k $$

are supposed to generate the exterior algebra. Your elements however will at most generate a subspace of $A$. This won't be isomorphic to the exterior algebra due to the problems with dimension. Of course there are some trivial cases where they do end up aligning.