Inducing maps on the exterior products

Question

Let $V, W$ be two vector spaces over a field $F$. It is known that if $f \colon V \rightarrow W$ is a linear transformation we can induce a linear map on the exterior products $\Lambda^kf \colon \Lambda^k V \rightarrow \Lambda^k W$ by just setting $\Lambda^kf(v_1\wedge \dots \wedge v_k)=f(v_1)\wedge \dots \wedge f(v_k)$.

But my question is: if we had $k$ linear maps $f_1, \dots, f_k\colon V \rightarrow W$ can we induce a linear transformation $\Lambda^k V \rightarrow \Lambda^k W$ involving all of them?

Let's take the easy case $k=2$ with maps $f=f_1$ and $g=f_2$. The most naïve idea would be to define $f\wedge g\colon v_1 \wedge v_2 \mapsto f(v_1)\wedge g(v_2)$ but it's immediate to see it is not well defined. Then I though of setting $f \wedge g-g \wedge f\colon v_1 \wedge v_2\mapsto f(v_1)\wedge g(v_2)-g(v_1)\wedge f(v_1)$ but with this is easy to see $v_1 \wedge v_2$ and $v_2 \wedge v_1$ are mapped to the same values while they should be opposite.

The fact is that I am dealing with formulas involving linear combinations of $\Lambda^k f_i$ for various $f_i$'s and I would like to express them in a nice way. For example take $V=W=F^n$ so the $f_i$'s are just $n \times n$ matrices. When $n=2$ I have to study the formula \begin{equation} \frac{1}{2}\biggl(tr\Lambda^2(f_1+f_2)-tr\Lambda^2f_1-tr\Lambda^2 f_2 \biggr). \end{equation}

To be rigorous you could answer that if I remove the trace from this formula I get a linear transformation $\Lambda^2 V \rightarrow \Lambda^2W$ involving $f_1$ and $f_2$ as I asked. But I am interested in a way to make such formulas more nice.

Answer 1

One way to do it is to define

$$ (f_1 \wedge \dots \wedge f_k)(v_1 \wedge \dots \wedge v_k) := \sum_{\sigma \in S_k} (-1)^{\sigma} f_1(v_{\sigma(1)}) \wedge \dots \wedge f_k(v_{\sigma(k)}). $$

You can check directly that this is well-defined and that $\underbrace{f \wedge \dots \wedge f}_{k \textrm{ times}} = k! \cdot \Lambda^k(f)$. For $k = 2$, you get

$$ (f \wedge g)(v_1 \wedge v_2) = f(v_1) \wedge g(v_2) - f(v_2) \wedge g(v_1). $$

Then

$$ 2 \cdot \Lambda^2(f_1 + f_2) = (f_1 + f_2) \wedge (f_1 + f_2) = f_1 \wedge f_1 + 2 f_1 \wedge f_2 + f_2 \wedge f_2 \\= 2 \left( \Lambda^2(f_1) + f_1 \wedge f_2 + \Lambda^2(f_2) \right)$$

so

$$ \Lambda^2(f_1 + f_2) - \Lambda^2(f_1) - \Lambda^2(f_2) = f_1 \wedge f_2 $$

and your expression is just half the trace of $f_1 \wedge f_2$.

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Remark: This might seem as an ad hoc definition but it actually quite natural from a certain perspective. Assuming $V,W$ are finite dimensional, we have $\operatorname{Hom}(\Lambda(V), \Lambda(W)) \cong \Lambda(V^{*}) \otimes \Lambda(W)$. Both $\Lambda(V^{*})$ and $\Lambda(W)$ are graded algebras so the tensor product inherits a natural multiplication defined by

$$ (\mu_1 \otimes \eta_1) \wedge (\mu_2 \otimes \eta_2) := (\mu_1 \wedge \mu_2) \otimes (\eta_1 \wedge \eta_2), \,\,\, \mu_i \in \Lambda(V^{*}), \eta_i \in \Lambda(W). $$

The resulting bi-graded algebra is called sometimes the mixed exterior algebra. It has inside a copy of $\Lambda(V^{*})$ and of $\Lambda(W)$. If you identify maps $f,g \colon V \rightarrow W$ as $(1,1)$ elements of the mixed exterior algebra, take their product and identify the resulting $(2,2)$ element with a map from $\Lambda^2(V)$ to $\Lambda^2(W)$, you get the definition I gave in the beginning of my answer.