What is the inverse of the isomorphism $\displaystyle \bigoplus_{j=0}^k \Lambda^{k-j}(E)\otimes \Lambda^j(F)\longrightarrow \Lambda^k(E\oplus F)$?

Question

Let $E$ and $F$ be two vector spaces. I want to show $$\Lambda^k(E\oplus F)\simeq \bigoplus_{j=0}^k \Lambda^{k-j}(E)\otimes \Lambda^j(F).$$ I thought doing something like: We have a linear map $$\Phi:\Lambda^{k-j}(E)\otimes \Lambda^j(F)\longrightarrow \Lambda^k(E\oplus F), \omega_1\otimes\omega_2\longmapsto \Lambda^{k-j}(\jmath_1)(\omega_1)\wedge \Lambda^j(\jmath_2)(\omega_2),$$ where

• $\jmath_1:E\longrightarrow E\oplus F$ and $\jmath_2:F\longrightarrow E\oplus F$ are the canonical injections;

• $\Lambda^{k-j}(\jmath_1):\Lambda^{k-j}(E)\longrightarrow \Lambda^{k-j}(E\oplus F)$ and $\Lambda^j(\jmath_2): \Lambda^{j}(F)\longrightarrow \Lambda^{j} (E\oplus F)$ are the induced maps on the exterior power by $\jmath_1$ and $\jmath_2$.

By the universal property of direct sums there is a unique linear map $$f:\bigoplus_{j=1}^k\Lambda^{k-j}(E)\otimes \Lambda^j(F)\longrightarrow \Lambda^k(E\oplus F),$$ such that $\Phi\circ \jmath=f$ where $$\jmath:\Lambda^{k-j}(E)\otimes \Lambda^j(F)\longrightarrow \bigoplus_{j=0}^k \Lambda^{k-j}(E)\otimes \Lambda^j(F)$$ is the cannonical injection. The map $f$ must be the required isomorphism.

How to define the inverse of $f$?

Maybe I could use some other argument to show $f$ is an isomorphism but I need the expression of the inverse for doing some computations.

Thanks.

Answer 1

Assume for simplicity of notation that $E,F$ are finite dimensional. Choose a basis $e_1, \dots, e_n$ for $E$ and $f_1, \dots, f_m$ for $F$. Then $(\jmath_1(e_1), \dots, \jmath_1(e_n), \jmath_2(f_1), \dots, \jmath_2(f_m))$ is a basis of $E \oplus F$ and using this basis we can construct a basis for $\Lambda^k(E \oplus F)$ that consists of elements of the form $(\Lambda^{k-j}(\jmath_1))(e_{I}) \wedge (\Lambda^k(\jmath_2))(f_{K})$ where:

1. The index $j$ ranges in $0 \leq j \leq k$.
2. $I = (i_1, \dots, i_{k-j})$ is an increasing enumeration of (some of) the elements of $1,\dots,n$ with $i_1 < \dots < i_{k - j}$ and $e_I = e_{i_1} \wedge \dots \wedge e_{i_{k-j}}$.
3. $K = (k_1, \dots, k_j)$ is an increasing enumeration of (some of) the elements of $1,\dots,m$ with $k_1 < \dots < k_j$ and $0 \leq j \leq k$ and $f_K = f_{k_1} \wedge \dots \wedge f_{k_j}$.

The inverse map will send $(\Lambda^{k-j}(\jmath_1))(e_{I}) \wedge (\Lambda^k(\jmath_2))(f_{K})$ to $e_I \otimes f_K$.