# Inducing maps on the exterior products

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 $$$$\frac{1}{2}\biggl(tr\Lambda^2(f_1+f_2)-tr\Lambda^2f_1-tr\Lambda^2 f_2 \biggr).$$$$

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.

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 $$f_1 \wedge f_2$$.

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 $$\Lambda(W)$$. 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.

• Are you sure that you do not need a factor $1/k!$ in the definition of $f_1 \wedge \dots \wedge f_k$? (We would need $F$ of characteristic $0$ obviously). If $W=F$ the base field then this should reduce to the usual wedge product of tensors and for this product we need the factor $1/k!$ to make the computations work (I think you need it for associativity?).
– N.B.
Oct 1, 2020 at 20:59
• N.B: You don't need it but you can add the factor if you want. It is already associative as it is. BTW, I'm using the pairing between $\Lambda^k(V^{*})$ with $\Lambda^k(V)$ given by $(\varphi^1 \wedge \dots \wedge \varphi^k)(v_1 \wedge \dots \wedge v_k) = \det(\varphi^i(v_j))$ without a $\frac{1}{k!}$ factor so it indeed generalizes the usual wedge product. Oct 1, 2020 at 22:01