Lemma 11.9(page 127)showed the relation between diagonal class $u''\in H^n(M\times M)$ and fundamental class $\mu\in H_n(M)$ for a compact manifold $M$ with dimention n: $$u''/\mu=1$$

where $u'':=u'|_{M\times M}$, $u'\in H^n(M\times M,M\times M-\Delta(M))$

By Tubular Neighborhood Theorem, if n-manifold $M$ is embedded in Riemannian manifold $A$ as a closed subsets, $H^n(E,E_0)$ of normal bundle is isomorphic to $H^n(A,A-M)$. By diagonal embedding $\Delta: M \rightarrow M\times M \quad x \mapsto (x,x)$, tangent bundle is diffeomorphic to the normal bundle associated with diagonal embedding. Thus, $H^n(TM,TM-M)\cong H^n(M\times M, M\times M -\Delta(M))$

The proof makes use of the commutative diagram enter image description here

where $/$ is slant product and $j_x: M \rightarrow M\times M \quad y\mapsto (x,y)$

Hence, we have $u''/\mu|_x=\langle j_x^*(u''),\mu\rangle=\langle j_x^*(u')|_M,\mu\rangle=\langle j_x^*(u'),\mu_x\rangle=\langle u_x,\mu_x\rangle=1$

$j_x^*(u')=u_x\in H^n(M,M-{x})$ is proved in lemma 11.7. $\mu_x \in H_n(M,M-\{x\})$.

Here i don't understand why $\langle j_x^*(u')|_M,\mu\rangle=\langle j_x^*(u'),\mu_x\rangle$. They mentioned it follows by defining property of $\mu$ (i am not quite familiar with fundamental class, so i ask here)

Another is the proof in Wu's Theorem(page 133). At the end of proof, it says $$Sq(v)=\sum Sq(b_i)\times Sq(b_i^{\#})/\mu=Sq(u'')/\mu$$ where Sq is the total Steenrod square and $/$ is slant product.

To me, $u''=\sum (-1)^{|b_i|}b_i\times b_i^{\#}$ (proved by duality theorem where $b_i$ is a basis for $H^*(M)$, $b_i^\#$ is a dual basis corresponding to $b_i$) and $\sum Sq(b_i)\times Sq(b_i^{\#})=Sq(\sum b_i\times b_i^{\#})$. It seems that the equation is not correct?

  • $\begingroup$ I am sorry I don't understand your question about Steenrod squares. Why do you think that the equation is not correct? $\endgroup$
    – feynhat
    Jun 24, 2023 at 23:28
  • $\begingroup$ i think the factor $(-1)^{|b_i|}$ missing? Did i make some mistake ? $\endgroup$
    – Jino
    Jun 24, 2023 at 23:33
  • 2
    $\begingroup$ Steenrod squares and SW classes are in mod-2 cohomology, so $\pm 1$ doesn't matter. $\endgroup$
    – feynhat
    Jun 24, 2023 at 23:36
  • $\begingroup$ oh, oh, thanks. $\endgroup$
    – Jino
    Jun 24, 2023 at 23:38

1 Answer 1


Observe the following about Kronecker index. Let $f: X \to Y$ be a map then we have an induced maps on homology, $f_*: H_n(X) \to H_n(Y)$ and on cohomology, $f^*: H^n(Y) \to H^n(X)$. Suppose $\sigma \in H_n(X)$ and $c \in H^n(Y)$, then, $$ \langle c, f_*\sigma \rangle = \langle f^*c, \sigma \rangle. $$ (This just follows from the definition of Kronecker index).

Now in our case we have the inclusion map $\rho_x : (M, \varnothing) \to (M, M-x)$. The map induced in cohomology $\rho_x^*: H^n(M, M-x) \to H^n(M)$ is just the restriction map (that is, it takes a cochain in $(M, M-x)$ and restricts it to chains in $(M, \varnothing)$).

Notice that $j_x^*(u') \in H^n(M, M-x)$ and $j_x^*(u')|_M$ is just $\rho_x^*(j_x^*(u')) \in H^n(M)$.

Likewise, $\mu_x = \rho_{x*}(\mu) \in H_n(M, M-x)$.


$$ \langle j_x^*(u')|_M, \mu\rangle = \langle \rho_x^*(j_x^*(u')), \mu \rangle = \langle j_x^*(u'), \rho_{x*}(\mu) \rangle = \langle j_x(u'), \mu_x \rangle, $$ where the middle equality comes from the result stated above.


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