# Does the Gysin map in $K$-theory respect bordism?

Let $$X_1$$ and $$X_2$$ be two closed spin$$^c$$ manifolds that are bordant via a spin$$^c$$ manifold-with-boundary $$W$$.

Let $$Z$$ be a closed spin$$^c$$ manifold with $$\dim Z=\dim X_1$$ mod $$2$$. Let $$f_1:X_1\to Z,\qquad f_2:X_2\to Z,\qquad F:W\to Z$$ be smooth maps such that $$F|_{X_1}=f_1$$ and $$F|_{X_2}=f_2$$. We can associate to $$f_1$$ and $$f_2$$ two wrong-way (or Gysin) maps in $$K$$-theory:

$$f_{1!}:K^0(X_1)\to K^0(Z),$$ $$f_{2!}:K^0(X_2)\to K^0(Z).$$

Let $$E_1\to X_1$$ and $$E_2\to X_2$$ be two $$\mathbb{C}$$-vector bundles such that there exists a vector bundle $$\Omega\to W$$ satisfying $$\Omega|_{X_1}\cong E_1$$ and $$\Omega|_{X_2}\cong E_2$$. Let $$[E_i]\in K^0(X_i)$$ denote the $$K$$-theory classes defined by $$E_i$$.

Question: Is it true that $$f_{1!}[E_1]=f_{2!}[E_2]\in K^0(Z)$$?

Added after: I would be most interested in an approach not directly using Poincare duality for K-theory/K-homology.

• I know how to do this in K-homology, which gives me confidence that it also works for K-theory, but I haven't worked it out. In K-homology you can attach cylindrical ends $X_1 \times [0, \infty)$ and $X_2 \times [0, \infty)$ to $W$ and argue that the Gysin maps factor through the Mayer-Vietoris boundary maps on the new object $W'$. You then lift the K-homology class on W to a class in a certain K-homology-like group for W'; this requires care since W' is non-compact. (One way to do it involves C*-algebras and coarse geometry.) Functoriality of Mayer-Vietoris completes the proof. Aug 16, 2020 at 17:40
• @PaulSiegel Could you elaborate how the Gysin maps factor through the Mayer-Vietoris boundary maps on $W'$ and why this implies the result in $K$-homology? I would like to adapt your argument to $K$-theory. Aug 18, 2020 at 2:11

Let $$N^n=\partial M^{n+1}$$, $$E\in K^\bullet(M)$$ and $$f:M\to X$$

Choose a smooth embedding $$i:X\to \mathbb{R}^N,N>>1$$, denote by $$\chi$$ the normal bundle of $$X$$ and by $$\mu$$ the normal bundle of $$M$$ after suitable small deformation of $$i\circ f$$.

Let $$\nu=\mu|_N$$ and $$\eta$$ be the normal bundle of $$N\subset M$$ (which is trivial and one-dimensional)

By considering tubular neighborhoods we get the natural map:

$$t:Th_\chi X\to Th_{\nu+\eta}N$$, where $$Th$$ denotes a Thom space.

After applying the Thom isomorphism $$th$$ on $$K^\bullet$$ we obtain the definition of a Gysin map (going in "right-way" on a $$Th$$'s). So for $$f_!(E|_N)=0$$ it's sufficient to prove that $$t^* th_{\nu+\eta}(E|_N)=0$$

Actually $$t^*$$ is passing through a connecting homomorphism. Namely, there is a commutative diagram:

$$\begin{matrix} Th_{\chi}X&\to& Th_{\mu}M/Th_\nu N&\\ \downarrow{t}&\swarrow{\sigma}&\downarrow{\Sigma}&\\ Th_{\nu+\eta}N&\xrightarrow{\sim}& \Sigma Th_{\nu}N&\\ \end{matrix}$$

The top arrow comes from the tubular neighborhoods.

The horizontal isomorphism comes from triviality of $$\eta$$, while suspension $$\Sigma$$ from Puppe cofiber sequence:

$$Th_\nu N\to Th_\mu M\to Th_\mu M/Th_\nu N\xrightarrow{\Sigma} \Sigma Th_{\nu}N$$

The map $$\sigma$$ explains commutativity and is coming from:

$$Th_\mu M/Th_\nu N\sim Th_\mu M/Th_\mu (N\times [0,\varepsilon))\to$$ $$Th_\mu (N\times(-\varepsilon,\varepsilon))/Th_\mu (N\times [0,\varepsilon))\to Th_{\nu+\eta}N$$ where $$N\times [0,\varepsilon)\subset M$$ is a collar of $$N$$.

Finally, $$\Sigma^*$$ is the connecting homorphism and it follows that $$\Sigma^* th_{\nu}(F)=0$$ for all $$F\in Im( K^\bullet(M)\to K^\bullet(N))$$, so $$t^* th_{\nu+\eta}(E|_N)=0$$

• I have a couple of questions. How is the map $t$ defined? Also, it seems to me that $\nu\oplus\eta$ is just the normal bundle of $N$ in $\mathbb{R}^N$ - is this correct? Aug 21, 2020 at 8:30
• @geometricK, correct. Let $D$ and $S$ denote a small disk and sphere bundles respectively. I assume that $D_{\nu+\eta}N\subset D_\chi X$. This inclusion automatically gives a "wrong-way" map $t:D_\chi/S_\chi\to D_{\nu+\eta}/S_{\nu+\eta}$ The horizontal map is obtained in the same way Aug 21, 2020 at 20:16
• Am I correct in understanding that your argument in fact shows a stronger result, namely that the Gysin map $i_!$ for the inclusion $i:\partial M\hookrightarrow M$ is always $0$? (I assume $i_!$ is the $K$-map induced by the composition $Th_{\mu}M\to Th_{\mu}M/Th_{\nu}N\xrightarrow{\Sigma}\Sigma Th_{\nu}N\sim Th_{\nu+\eta}N)$. Then the fact that $f_!$ factors through $i_!$ gives the result I was after originally. Aug 22, 2020 at 2:12
• @geometricK, of course we can forget about $X$ and prove that $i_!(E|_N)$ is zero. Always in a sense that we have to know that $\Sigma^*(E_N)=0$, which is coming from $K^\bullet(M/N\xrightarrow{\Sigma} \Sigma N\to \Sigma M)$, then the same holds after applying $th$-isomorphism, so $\Sigma^* th_\nu (E|_N)=0$. I don't know how to prove that the two long sequences naturally isomorphic at $K^\bullet(M/N)$ and $K^\bullet(Th_\mu(M)/Th_\mu(N))$, but for required vanishing it's not essential Aug 22, 2020 at 8:47
• Thanks. There should be a relative version of the Thom isomorphism, which I'm trying to locate and which should give the isomorphism $K^*(M/N)\to K^*(Th_{\mu}M/Th_{\nu}N)$. As you say this isn't strictly necessary but would still be nice to have. Aug 25, 2020 at 6:29

The answer is yes, using general properties of orientations and fundamental classes.

Let $$X_1$$ and $$X_2$$ be $$n$$--dimensional. Then $$f_{!i}$$ is the composite $$K^0(X_i) \xrightarrow[\sim]{\cap [X_i]} K_n(X_i) \xrightarrow{f_{i*}} K_n(Z) \xleftarrow[\sim]{\cap [Z]} K^0(Z).$$

Meanwhile Poincare duality for $$W$$ has the form $$K^0(W) \xrightarrow{\cap [W]} K_{n+1}(W, X_1 \coprod X_2)$$, and $$d([W]) = [X_1]-[X_2]$$. Thus $$d(\Omega \cap [W]) = (E_1 \cap [X_1], -E_2 \cap [X_2])$$, and so

$$(f_{1*})(E_1 \cap [X_1]) - (f_{2*}(E_2 \cap [X_2]) = F_* i_* (d(\Omega \cap [W])) = 0,$$

since the composite

$$K_{n+1}(W,X_1\coprod X_2) \xrightarrow{d} K_n(X_1 \coprod X_2) \xrightarrow{i_*} K_n(W)$$

is zero.

• Although I completely agree with your answer, I was secretly hoping for an approach without Poincare duality, as (although I didn't mention this) Poincare duality may not be available to me in the setting I'm actually working in. Aug 16, 2020 at 20:58