According to the Wikipedia article on the exterior algebra, there is a natural isomorphism
$$\bigwedge(V\oplus W) \simeq \bigwedge V \otimes \bigwedge W,$$
where $V,W$ is a finite-dimensional $\Bbbk$-vector space.
My question is, in what sense they are isomorphic? Isomorphic as a vector space? (this is trivial.) Or isomorphic as a $\Bbbk$-algebra?
I hope that they are isomorphic as a $\Bbbk$-algebra, but I think this is not true due to the anticommutative nature. In particular, for $v_1,v_2\in V$ and $w_1,w_2\in W$,
$$(v_1\otimes w_1)(v_2\otimes w_2) = (v_1\wedge v_2) \otimes (w_1\wedge w_2)$$
in RHS but it is not true in LHS that
$$v_1\wedge w_1 \wedge v_2 \wedge w_2 = v_1\wedge v_2\wedge w_1\wedge w_2.$$
(However, for symmetric algebra, analogous relation is true and the isomorphism is in the $\Bbbk$-algebra sense.)
Is my reasoning correct?
Best Answer
Your reasoning is not necessarily incorrect, but the issue you see can be mitigated. To do this, we have to slightly ponder what "kind of object" exterior algebras are and what the symbol $\otimes$ means. Namely, if $V$ is a $k$-vector space, then $\bigwedge V$ is not just a $k$-algebra, but a graded $k$-algebra. In fact, it is a graded commutative $k$-algebra (the Wikipedia article calls this "anticommutative" instead - I personally disagree with that choice of terminology), i.e. if $\omega,\eta\in\bigwedge V$ are elements of pure degree, then $\omega\cdot\eta=(-1)^{|\omega||\eta|}\eta\cdot\omega$ (the multiplication here is the exterior product, I use bars to denote the degree).
Now, in general, if you have two graded $k$-algebras $R=\bigoplus_{i\ge0}R_i$ and $S=\bigoplus_{j\ge0}S_j$, then their tensor product $R\otimes_kS=\bigoplus_{k\ge0}\bigoplus_{i+j=k}R_i\otimes_kS_j$ is naturally a graded $k$-vector space with the indicated grading. In fact, $R\otimes_kS$ can be naturally made into a graded $k$-algebra itself, but here is the surprise. If $r\in R_i,s\in S_j,r^{\prime}\in R_{i^{\prime}}$ and $s^{\prime}\in S_{j^{\prime}}$, we ought to define $(r\otimes s)\cdot(r^{\prime}\otimes s^{\prime})=(-1)^{ji^{\prime}}(r\cdot r^{\prime})\otimes(s\cdot s^{\prime})$. This is in accordance with a general principle known as the "Koszul sign rule", which says that when you exchange two graded symbols of degree $j$ and $i^{\prime}$ ($s$ and $r^{\prime}$ here), the sign $(-1)^{ji^{\prime}}$ ought to be inserted. This makes $R\otimes_kS$ into a graded $k$-algebra, which we call the graded tensor product of $R$ and $S$.
Here's a concrete reason why this is the "right" definition: If $R$ and $S$ are graded commutative, then $R\otimes_kS$ is graded commutative too and it would not be if you omitted the sign (check this). Next, if $T$ is any graded commutative $k$-algebra, the two natural $k$-algebra homomorphisms $R\rightarrow R\otimes_kS,\,r\mapsto r\otimes1$ and $S\rightarrow R\otimes_kS,\,s\mapsto1\otimes s$ induce a bijection between pairs of graded $k$-algebra homomorphisms $R\rightarrow T$ and $S\rightarrow T$ and $k$-algebra homomorphisms $R\otimes_kS\rightarrow T$ in the way one would expect (check this and you'll see the sign is crucial here as well). In other words, the graded tensor product is the coproduct in the category of graded commutative $k$-algebras. This generalizes how the (ungraded) tensor product works to the graded setting.
If we now return to the scenario at hand, this deliberation tells us that $\bigwedge V\otimes_k\bigwedge W$ ought to be understood not as a tensor product of algebras, but as a graded tensor product of graded $k$-algebras. The additional sign in the definition of the multiplication in a graded tensor product is precisely the sign that is missing in the equations you write down, so the natural isomorphism $\bigwedge V\otimes_k\bigwedge W\rightarrow\bigwedge(V\oplus W)$ does in fact become an isomorphism of graded commutative $k$-algebras.
Of course, I could've just directly told you to add a sign to the multiplication in $\bigwedge V\otimes_k\bigwedge W$ that magically makes everything work, but I hope this explanation, even if it may not be entirely transparent, highlights that this is not some ad hoc fix, but fundamentally the right perspective on the issue.