3

There is an argument that I would like to fully understand, but I can't see it, yet.

Here is the situation: Given smooth manifolds $X$, $Y$ and $Z$ and transverse maps $f:X\rightarrow Z$, $g\rightarrow Z$, then the fibered product $M:=X\times_{f=g} Y$ is well defined. In fact, it is given by points $(x,y)\in X\times Y$ such that $f(x) = g(y)$. In other words, it is the preiamge of the diagonal $\Delta_Z\subset Z\times Z$ of the map $F=(f,g): X\times Y\rightarrow Z\times Z$.

I am looking for an expression for the tangent bundle $TM$ using the language of $K$-theory (i.e. a virtual bundle). I have read the following argument: The normal bundle of $\Delta_Z$ in $Z\times Z$ is isomorphic to (the pullback of) $TZ$, so it follows that the tangent bundle of the fibered square is $TM = TX+TY - f^*TZ$.

It must have something to do with the transversallity of $f$ and $g$ but I can't fill in the details. Any help is appreciated:)

Teddyboer
  • 748
  • Yes, it's transversality. Perhaps you should warm up with the more general result: If $f\colon X\to Y$ is smooth and is transverse to $Z\subset Y$, then $W=f^{-1}(Z)$ is a submanifold and $N(W,X) = f^*N(Z,Y)$. The definition of transversality tells you that $df_p$ maps $N_p(W,X)$ isomorphically to $N_{f(p)}(Z,Y)$. – Ted Shifrin May 09 '23 at 17:41
  • So I am very much aware of the first part, i.e. that if $f$ is transverse to $Z$, then $f^{-1}(Z)$ is a submanifold of codimension $\text{cod}(Z)$. It uses a neat little fact from linear algebra that if we have a sequence of linear maps $A\overset{\alpha}{\rightarrow} B \overset{\beta}{\rightarrow} C$ with $\beta$ surfective, then $B = \text{im}(\alpha) + \text{ker}(\beta)$ if and only if $\beta\circ \alpha$ is surjective. I havent't looked at the normal bundle, yet. It should follows similarly. Thanks for the hint. – Teddyboer May 09 '23 at 17:47
  • Yes, I like the viewpoint of that neat little fact. I have mentioned it in teaching, but don’t remember seeing it in the standard texts. – Ted Shifrin May 09 '23 at 18:02
  • My supervisor used this argument in his lecture on differential topology. But it can also be found on page 52 of Golubitskky's and Guillemin's "stable mappings and their singularities". It's a very local thinking. – Teddyboer May 09 '23 at 18:48
  • As for the second statement, I think I get the idea. First, linear algebra: Take $\alpha: V\rightarrow W$ linear and $U\subset W$ a subspace with $\text{im}(\alpha)+U = W$. Then it is easy to show that $W/U\cong V/\alpha^{-1}(U)$. I.e. the complementary space of $U$ in $W$ is isomorphic to the complementary space of $\alpha^{-1}(U)$ in $V$. In this situation, we have $M =f^{-1}(Z)$ so let $\alpha = df(x): V:=T_xX\rightarrow W:=T_{f(x)}Y$, $U = T_{f(x)}Z$. Transversality shows that $\text{im}(\alpha) + U = df(x)(T_x X) + T_{f(x)}Z = T_{f(x)}Y$. – Teddyboer May 10 '23 at 08:56
  • Moreover, it is known that $T_x M = (df(x))^{-1}(T_{f(x)}Z)$. Then the isomorphism from linear algebra shows $T_{f(x)}/T_{f(x)}Z \cong T_x X/ T_xM$ for all $x\in M$. So we have an ismorphism on the fibers of $N(M,X)$ and $f^N(Z,Y)$ which induces a vector bundle isomorphism $df: N(M,X)\rightarrow f^N(Z,Y)$ – Teddyboer May 10 '23 at 09:03
  • As for my main question: The transversality of $f$ and $g$ implies that $F=(f,g)$ is transverse to the diagonal $\Delta_Z$. Thus, $M=F^{-1}(\Delta_Z)$ is indeed a submanifold and the normal bundle $N(M,X\times X)$ is isomorphic to $F^N(\Delta_Z,Z\times Z)$. Denote by $\pi_1: Z\times Z \rightarrow Z$ the projection onto the first factor. The normal bundle $N(\Delta_Z,Z\times Z)$ is isomorphic to $\pi_1^TZ$, thus $F^N(\Delta_Z,Z\times Z) \cong f^TZ$. – Teddyboer May 10 '23 at 10:02
  • Looks good. You should put this all into an answer. – Ted Shifrin May 10 '23 at 20:45

1 Answers1

1

With some help by Ted Shifrin, I was able to find a solution:

First let $f:X\rightarrow Y$ be smooth, $Z\subset Y$ and $f$ transverse to $Z$. Then it is a well known fact that $W = f^{-1}(Z)$ is a submanifold.

I want to show that $N(W,X) \cong f^*N(Z,Y)$ by defining a suitable bundle isomorphism:

Let $x\in X$ be such that $f(x)\in Z$. It is known see here that $T_xW = (df(x))^{-1}(T_{f(x)}Z)$. Consider the induced map $g: T_x X\rightarrow T_{f(x)}Y/T_{f(x)}Z$, $v\mapsto df(x)[v]+T_{f(x)}Z$. The kernel of this map is $$\text{ker}(g) = (df(x))^{-1}(T_{f(x)}Z) = T_xW.$$ By transversality, we find the image of $g$ to be $$ \text{im}(g) = (df(x)(T_xX) + T_{f(x)}Z)/T_{f(x)}Z = T_{f(x)}Y/T_{f(x)}Z,$$ so we obtain an isomorphism $T_{x}X/T_xW\overset{\sim}{\longrightarrow} T_{f(x)}Y/T_{f(x)}Z$ induced by $df(x)$. This map induces a bundle isomorphism $N(X,W) = TX/TW \cong f^* TY/TZ = f^*N(Z,Y)$.

For the original question: Let $M = F^{-1}(\Delta_Z)$. Denote by $\pi_i:Z\times Z\rightarrow Z$ the projections onto the first and second factor, respectively. The kernel of $d\pi_2$ is exactly $\pi_2^* TZ$ and the restriction of $d\pi_2$ to $T\Delta_Z$ defines an isomprhism to $\pi_2^*TZ$. So $\pi_1^*TZ$ is isomorphic to the normal bundle of $T\Delta_Z$ in $Z\times Z$.

Let $(x,y)$ be such that $f(x) = g(y)$, then transversality gives $$dF(x,y)(T_{(x,y)}X\times Y) + T_{F(x,y)}\Delta_Z = df(x)(T_xX) + dg(y)(T_yY) + T_{F(x,y)}\Delta_Z = T_{f(x)}Z + T_{g(y)}Z = T_{F(x,y)}(Z\times Z),$$ so $F$ is transverse to $\Delta_Z$. Thus, $M$ is indeed a submanifold of $X\times Y$ and $$ T(X\times Y)/TM = N(M,X\times Y) \cong F^*N(\Delta_Z,Z\times Z) \cong F^*\pi_1^* TZ.$$ Since $\pi_1\circ F\circ j_X = f$, where $j_X:X\hookrightarrow X\times Y$, we can write $$ TM = TX + TY - f^*TZ.$$

There might be some minor technical issues but the idea should (hopefully) be correct.

Teddyboer
  • 748