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I am trying to learn about Thom polynomials and I often find arguments in the literature that just make no real sense to me. Maybe it is due to my lack of knowledge in $K$ - theory. I apologize for the lengthy post, but maybe someone knows these things well.

Let $P := \mathbb{C}P^n$, $N := \mathbb{C}P^{n*}$ and set $X := P\times N$. Denote by $p_1:X\rightarrow P$ and $p_2: X\rightarrow N$ the natural projections. Let $V = \mathbb{V}(F)\subset P$ be a hypersurface of degree $d$ and $H = \{(x,a^*)\in P\times N:\text{ }a^*(x) = 0\}$ be the incidence variety of points and lines in $P$. In fact, $H$ can be realized as the zero-set (zero-scheme) of a section of the line bundle $p_1^*\mathcal{O}_P(1)\otimes p_2^*\mathcal{O}_N(1)$ over $X$.

Also, the polynomial $F$ defines a section of the line bundle $p_1^*\mathcal{O}_P(d)$ over $X$. Hence, there is an induced section of the rank $2$ vector bundle $E =p_1^*\mathcal{O}_P(d)\oplus (p_1^*\mathcal{O}_P(1)\otimes p_2^*\mathcal{O}_N(1))$ over $X$. Its zero-set (zero-scheme) is the subvariety $M :=\{ (x,a^*)\in V\times N:\text{ }a^*(x) =0\}$, i.e. the incidence variety of points in $V$ and lines in $P$.

Define $f:M\rightarrow N$ by composing $p_2:X\rightarrow N$ with the inclusion $i:M\hookrightarrow X$, i.e. $f = p_2\circ i$. I am interested in computing the total Chern class $c(f^*TN-TM)$, where $f^*TM-TN$ is the virtual bundle living in the $K$ - group of $M$.

The claim is: $$TM = i^*(TX-E)\text{ and hence } f^{*}TN - TM = i^{*}(E - p_1^*TP).$$

It appears to me that these are just standard calculations in $K$ - groups but I don't get it.

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I now have some idea on how to prove these relations. Consider a real vector bundle $\pi:E\rightarrow X$ of rank $k$, where $E$ and $X$ are $C^\infty$ manifolds and let $X$ be $n$,-,dimensional. If $s$ is a section transverse to the zero - section, then its zero set $Z(s) = \{x\in X|\text{ }s(x)=0_x\in E_x\}$ is a submanifold of $X$. What's more, on local bundle charts $U\subset X$, the restriction $s|U$ is a submersion in points $x\in U$ with $s(x) = 0$, meaning $\text{im}(ds(x))= \mathbb{R}^k$.
Let $M := Z(s)$, then the tangent space $T_xM$ can be identified with $\text{ker}(ds(x))$ and the normal space is isomorphic to $\mathbb{R}^k$, which follows from the canonical exact sequence $$0\rightarrow T_xM = \text{ker}(ds(x))\rightarrow T_xX\rightarrow N_xM = T_xX/T_xM\rightarrow 0.$$ Hence, $TX = TM \oplus i^*E$, where $i:M\hookrightarrow X$ is the inclusion. This proves $$TM = TX-i^*E\in K_0(X)$$ The second identity follows from $TX = TP\oplus TN$.

Although the computation above is valid for real vector bundles, I think we can interpret any complex vector bundle as a real vector bundle of twice the rank. Also $\mathbb{C}P^n$ is a $2n$ - dimensional real manifold...

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