# The proof of Lemma11.9 and theorem 11.14 (Wu) (Milnor & Stasheff)

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

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?

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

## 1 Answer

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

So,

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