Hatcher defines the cohomology ring $H^*(X;R)$ for a space $X$ and commutative ring $R$ to be the direct sum of the cohomology groups $H^n(X;R)$, and by his definition of a graded ring, elements of $H^n(X;R)$ will have dimension $n$.

In his discussion of the cross product $H^*(X;R)\otimes_R H^*(Y;R) \rightarrow H^*(X \times Y; R); a \otimes b \mapsto a \times b$ where $\otimes_R$ denotes the tensor product as $R$-modules.

He says that this becomes a ring homomorphism if we define the multiplication in a tensor product of graded rings by $(a \otimes b)(c \otimes d)=(-1)^{|b||c|}ac \otimes bd$ where $|x|$ denotes the dimension of $x$.

My problem here is that $a$ and $b$ are general elements of their respective cohomology rings, so are finite sum $a=\sum{\alpha_{n_i}}$ where each $\alpha_{n_i}$ is in some cohomology ring $H^{n_i}(X;R)$ and similarly for $b=\sum{\beta_{n_j}}$ For example, if $a=\alpha_1 +\alpha_2$ where $\alpha_i \in H^i(X;R)$, then what is the dimension of $a$ supposed to be?


You are right, a element $x \in H^*(X,R)$ does not have a fixed degree. But since multiplication should be bilinear, it's enough to define the multiplication for "pure" elements on the form $a \otimes b$ where $a,b$ are of fixed degree.

  • $\begingroup$ Just to add: sometimes you see "the definition extends by linearity." This is code for "we are using the universal property of tensor products given this bilinear map defined on the basis vectors." $\endgroup$ – Kyle Miller Jun 25 '17 at 1:22
  • $\begingroup$ So how exactly does it extend from pure elements? $\endgroup$ – TuoTuo Jun 25 '17 at 2:08
  • 1
    $\begingroup$ @TuoTuo : Write $a = \sum a_i$ where $a_i$ are of pure degree and similary for other terms. We obtain $$( (\sum a_i) \otimes (\sum b_j) )((\sum c_k) \otimes (\sum d_l)) = \sum_i (a_i \otimes(\sum b_j) \otimes ((\sum c_k) \otimes (\sum d_l)) = \dots = \sum_{i,j,k,l} (-1))^{jk} a_ic_k \otimes b_j d_l $$ and the last expression contains only products of pure elements. $\endgroup$ – user171326 Jun 25 '17 at 6:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.