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.

  • $\begingroup$ 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. $\endgroup$ Aug 16, 2020 at 17:40
  • $\begingroup$ @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. $\endgroup$
    – geometricK
    Aug 18, 2020 at 2:11

2 Answers 2


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 $t^* th_{\nu+\eta}(E|_N)=0$

  • $\begingroup$ 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? $\endgroup$
    – geometricK
    Aug 21, 2020 at 8:30
  • $\begingroup$ @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 $\endgroup$ Aug 21, 2020 at 20:16
  • $\begingroup$ 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. $\endgroup$
    – geometricK
    Aug 22, 2020 at 2:12
  • $\begingroup$ @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 $\endgroup$ Aug 22, 2020 at 8:47
  • $\begingroup$ 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. $\endgroup$
    – geometricK
    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.

  • $\begingroup$ 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. $\endgroup$
    – geometricK
    Aug 16, 2020 at 20:58

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