Does a Gysin map depend on the choice of Thom class?

Question

Let $f:X\rightarrow Y$ be a proper embedding between complex manifolds, then the normal bundle $N$ is complex which is in paticular $\textsf{spin}^c$. Hence we have a Thom class $\lambda_N$ and a Thom isomorphism $$K^{\bullet}(X)\rightarrow \tilde{K}^{\bullet}(\text{Th}(N))$$ in topological $K$-theory. When $X,Y$ are compact, we can then define a Gysin-Thom map $f_*:K^{\bullet}(X)\rightarrow K^{\bullet}(Y)$ by tubular neighbourhood theorem.

Does the Gysin map depend on our choice of the Thom class (aka. the $\textsf{spin}^c$ structure) on $N$?

Answer 1

Yes it does! The map in your display is (more or less) the cup product with the Thom class $\lambda_N$. So if you choose a different Thom class you'll get a different map.

For example, take the standard inclusion $\mathbb{CP}^n\hookrightarrow\mathbb{CP}^{n+1}$. The normal bundle is the tautological bundle $L\rightarrow \mathbb{CP}^n$ (or maybe its inverse, I forget). Now $K^0(\mathbb{CP}^n)\simeq\mathbb{Z}[x]/x^{n+1}$ where $x=1-[L]$. Up to a multiple of the Bott class $\beta$ we can choose the Thom class of $L$ to be $1-[L]$. Then the map $K^0(\mathbb{CP}^n)\rightarrow K^0(\mathbb{CP}^{n+1})$ that we're after is the map $\mathbb{Z}[x]/x^{n+1}\rightarrow \mathbb{Z}[x]/x^{n+2}$ given by multiplication by $x$. On the other hand we could choose the Thom class of $L$ to be $xu(x)$, where $u(x)$ is any unit $1+xp(x)\in \mathbb{Z}[[x]]$ (this statement is part of the theory of "complex oriented cohomology theories"), and so we'll get a different map.