Quick preface - this is problem 14-3 in Lee's Intro to Manifolds and a past homework problem I never really finished.

Let $\Re=$ space generated by all $w^1\otimes\cdots\otimes w^k|w^i=w^j$ for some $i\neq j$, and $A^k(V^*) = (V^*\otimes\cdots\otimes V^*)/\Re$

Define the wedge product on $\oplus_kA^k(V^*)$ by $\omega\wedge\eta=\pi(\tilde\omega\otimes\tilde\eta)$, where $\pi$ is the canonical projection from $(V^*\otimes\cdots\otimes V^*)$ to $A^k(V^*)$, and $\tilde\omega$ and $\tilde\eta$ are elements of $(V^*\otimes\cdots\otimes V^*)$ such that $\pi(\tilde\omega)=\omega$ and $\pi(\tilde\eta)=\eta$. Show that the wedge product as defined is well-defined.

I get the general idea is to show that $\omega\wedge\eta$ are determined regardless of choice of $\tilde\omega$ and $\tilde\eta$ (i.e. for $\tilde\omega_1\neq\tilde\omega_2, \tilde\eta_1\neq\tilde\eta_2$, we get $\pi(\tilde\omega_1\otimes\tilde\eta_1)=\pi(\tilde\omega_2\otimes\tilde\eta_2)$), but I'm not quite sure how to show it. I assume it's something to do with how the quotient and tensor product interact (and how that affects the projection mapping), but I haven't taken any category-theoretic class on this specific topic and was hoping for a more concrete answer. Any help is welcome!


1 Answer 1


Let $\mathfrak R^k$ be the subspace of $(V^*)^{\otimes k}$ generated by all the pure tensors $v^1 \otimes \cdots \otimes v^k$ such that $v^i = v^j$ for some distinct indices $i$ and $j$. Also, let $\pi^k$ be the canonical projection from $(V^*)^{\otimes k}$ onto $A^k(V^*) = (V^*)^{\otimes k}/\mathfrak R^k$.

Now, keep in mind that we want to prove that if $\tilde\omega^1$ and $\tilde\omega^2$ are tensors in some $(V^*)^{\otimes p}$ satisfying $\pi^p(\tilde\omega^1) = \pi^p(\tilde\omega^2)$, and if $\tilde\eta^1$ and $\tilde\eta^2$ are tensors in some other $(V^*)^{\otimes q}$ satisfying $\pi^q(\tilde\eta^1) = \pi^q(\tilde\eta^2)$, then $$\pi^{p+q}(\tilde\omega^1 \otimes \tilde\eta^1) = \pi^{p+q}(\tilde\omega^2 \otimes \tilde\eta^2). \tag{$*$}$$

To do this, note that $\pi^p(\tilde\omega^1) = \pi^p(\tilde\omega^2)$ means $\tilde\omega^1 - \tilde\omega^2 \in \mathfrak R^p$, and that $\pi^q(\tilde\eta^1) = \pi^q(\tilde\eta^2)$ means $\tilde\eta^1 - \tilde\eta^2 \in \mathfrak R^q$, so it follows that $$\tilde\omega^1 \otimes \tilde\eta^1 - \tilde\omega^2 \otimes \tilde\eta^2 = \tilde\omega^1 \otimes (\tilde\eta^1 - \tilde\eta^2) + (\tilde\omega^1 - \tilde\omega^2) \otimes \tilde\eta^2 \in \mathfrak R^{p+q}$$ (why?) and then we have $(*)$.


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