At the beginning of the proof(theorem 11.10 page 128), it says that it follows easily that the diagonal class can be expressed as r-fold sum \begin{align*} u''=b_i\times c_i+\ldots+b_r\times c_r \end{align*} by using the Kunneth formula \begin{align*} H^*(X\times Y)\cong H^*(X)\otimes H^*(Y) \end{align*} where $b_i,c_i$ are basis for $ H^*(M) $ and $|b_i|+|c_i|=n=\dim(M)$. $u'':=u'|_{M\times M}$ with

$u'\in H^{n}(M\times M,M\times M-\Delta(M))\cong H^{n}(TM,TM-M)$

Here the isomorphism is derived by the diagonal embedding $\Delta:M \rightarrow M\times M\quad x\mapsto (x,x)$ and Tubular Neighborhood Theorem

I think I miss something here, I only know that $H^n(M \times M)=\oplus_{i+j=n}H^{i}(M)\otimes H^{j}(M)$. So i don't know why $u''$ has this explicit form instead of $u''=b_1\times c_1$ or some other formula.

The answer below pointed out my flow. Thanks a lot.

  • $\begingroup$ What ring are you working with for your coefficients? Künneth gives you a short exact sequence $$0 \to H^∗(X,L) \otimes H^∗ (Y,M) \to H^∗ (X\times Y,L\otimes M) \to \text{Tor}^1 (H^∗ (X,L), H^∗ (Y,M)) \to 0$$ So you only have an iso in specific cases such as $L = M$ and cohomologies are finite dimensional. I assume here $L = M = \mathbb{Z}$. Ok I did not see that you had finite basis my bad :) $\endgroup$ Jun 24, 2023 at 14:26
  • $\begingroup$ I think this has to do with the fact that $u''$ is the diagonal class, that's why you have each coefficient with scalar $1$ $\endgroup$ Jun 24, 2023 at 14:33

1 Answer 1


They don't claim that $\{c_1, \dots, c_r\}$ is a basis for $H^*(M)$, they only assume $\{b_1, \dots, b_r\}$ is a basis for $H^*(M)$.

Any element of $H^*(X)\otimes H^*(Y)$ can be written as $\sum_{i=1}^pu_i\otimes v_i$ where $u_1, \dots, u_p \in H^*(X)$ and $v_1,\dots, v_p \in H^*(Y)$. The isomorphism $H^*(X)\otimes H^*(Y) \to H^*(X\times Y)$ is generated by $x\otimes y \mapsto \pi_1^*x\cup\pi_2^*y = x\times y$, so $\sum_{i=1}^pu_i\otimes v_i \mapsto \sum_{i=1}^pu_i\times v_i$. That is, every element of $H^*(X\times Y)$ can be written as $\sum_{i=1}^pu_i\times v_i$ for some $u_1, \dots, u_p \in H^*(X)$ and $v_1,\dots, v_p \in H^*(Y)$.

Now consider the case $X = Y = M$ and suppose $\{b_1, \dots, b_r\}$ is a basis for $H^*(M)$. Then for any $u_i \in H^*(M)$, we have $u_i = \sum_{j=1}^rm_{ij}b_j$ where $m_{ij} \in \Lambda$, the field of coefficients. Therefore

\begin{align*} \sum_{i=1}^pu_i\times v_i &= \sum_{i=1}^p\left(\sum_{j=1}^rm_{ij}b_j\right)\times v_i\\ &= \sum_{i=1}^p\sum_{j=1}^rm_{ij}b_j\times v_i\\ &= \sum_{j=1}^r\sum_{i=1}^pm_{ij}b_j\times v_i\\ &= \sum_{j=1}^r\sum_{i=1}^pb_j\times m_{ij}v_i\\ &= \sum_{j=1}^rb_j\times\left(\sum_{i=1}^pm_{ij}v_i\right)\\ &= \sum_{j=1}^rb_j\times c_j \end{align*}

where $c_j := \sum_{i=1}^pm_{ij}v_i$.

  • $\begingroup$ So at this moment we don't know if $c_j$ is zero or not? We know it at the end of the proof? $\endgroup$
    – Jino
    Jun 24, 2023 at 14:55
  • 1
    $\begingroup$ At this stage of the proof, they are just using the fact that the diagonal class, like any class in $H^*(M\times M)$, can be written as $b_1\times c_1 + \dots + b_r\times c_r$ for some $c_1, \dots, c_r \in H^*(M)$. Using the properties of the diagonal class, they then determine what the classes $c_i$ are. $\endgroup$ Jun 24, 2023 at 14:57
  • $\begingroup$ thanks. i assumed $c_j$ are also basis. $\endgroup$
    – Jino
    Jun 24, 2023 at 15:00
  • 1
    $\begingroup$ @Jino You do not need to assume that. The classes $c_i$'s are just undetermined elements in $H^*(M)$. In the course of the proof they determine that $c_j = (-1)^{\text{deg} b_j}b_j^\#$ but you do not and in fact, cannot assume that $c_j$'s form a basis because not every cohomology class in the product will be cross product of basis elements. $\endgroup$
    – feynhat
    Jun 24, 2023 at 23:04
  • $\begingroup$ yes, this is the reason i got confused $\endgroup$
    – Jino
    Jun 24, 2023 at 23:07

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