Homomorphism of Exterior power into Tensor power

Question

Suppose $\lambda (e_i \land e_j) = 1/2 * (e_i \otimes e_j - e_j \otimes e_i)$, with extension by linearity. I am not sure how to prove that $\lambda$ is an injective homomorphism because I am not able to work with the products the way I can with regular module and ring homomorphisms.

Answer 1

Lemma: If $f$ and $g$ are functions with the property that $f \circ g$ is the identity, then $g$ is injective and $f$ is surjective.

We can apply this lemma to this problem, by taking $\pi(v \otimes w) = v \wedge w$ (extended by linearity), and observing that $\pi \circ \lambda$ is the identity.

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Judging from comments, your problem isn't the injective part, but the homomorphism part. Homomorphisms can be characterized using the first isomorphism theorem:

• Any bilinear function $\varphi: V \times V \to W$ extends uniquely to a linear function $V \otimes V \to W$. (and conversely, all linear functions are of this form)
• Any linear function $\varphi : V \otimes V \to W$ satisfying $\varphi(x \otimes x) = 0$ descends to a unique linear function $\Lambda^2 V \to W$. (and conversely, all linear functions are of this form)

which can be merged into a single condition:

Theorem: For any bilinear function $\varphi : V \times V \to W$ satisfying $\varphi(x,x) = 0$, there is a unique linear function $\Lambda^2 V \to W$ that sends $v \wedge w \mapsto \varphi(v,w)$. Conversely, all linear functions $\Lambda^2 V \to W$ are of this form.

It's easy to check $$ \varphi(v, w) = \frac{1}{2}\left( v \otimes w - w \otimes v \right) $$ has the needed property to define a homomorphism.

Answer 2

You have $\lambda\colon \bigwedge^2 V\to V\otimes V$. An element of the domain can be written uniquely as $$ \omega=\sum_{i<j}a_{ij}(e_i\wedge e_j) $$ Suppose $\lambda(\omega)=0$, so $$ \lambda(\omega)=\frac{1}{2}\sum_{i<j}a_{ij}(e_i\otimes e_j-e_j\otimes e_i)=0 $$ Now recall that $\{e_i\otimes e_j: 1\le i\le n,1\le j\le n\}$ is linearly independent in $V\otimes V$.