# A formula for the second Chern class of the tensor product of a line bundle and vector bundle

If possible I would like someone to prove or suggest a place to see the proof of this relation:

$$c_2(V \otimes L)=c_2(V)+(r−1)c_1(V)c_1(L)+ {r \choose 2} c_1(L)^2$$

Here $$L$$ is the line bundle and $$r$$ is the rank of the vector bundle $$V$$. The reference that is mentioned in the link is not freely available...

What I need in reality is only to express this relation for the $$r=2$$ case, that is, to show that $$c_2(V \otimes L)=c_2(V)+c_1(V)c_1(L)+ c_1(L)^2$$.

I know the relation the relation $$c_1(V \otimes L)=rc_1(L) + c_1(V)$$ and that the total chern class satisfies $$c(E)= 1 + c_1(E) +... +c_n(E)$$, and that $$c(V \otimes L)=\prod_j (1 + c_1(L_j') + c_1(L))$$ if we assume that $$V= \oplus_{j=1}^r L_j'$$. However, this relations do not seem to suffice to prove what I want. Thanks in advance!

• Sorry but I have a couple of questions: Why is $c(1)c(2)=c_2(V)$ and why is $c_1(V)=c(1)+c(2)$? And also, when we write $c$ using these t's, they sort of serve to indicate what term corresponds to which $c_i$ right? @hm2020
– user770533
Dec 3, 2021 at 11:53

Question: "Sorry but I have a couple of questions: Why is c(1)c(2)=c2(V) and why is c1(V)=c(1)+c(2)? And also, when we write c using these t's, they sort of serve to indicate what term corresponds to which ci right? @hm2020"

Answer: when $$r=2$$ the formula follows from your relation $$c(V \otimes L)=\prod_j (1 + c_1(L_j') + c_1(L))$$: Let $$c(i):=c_1(L_i)$$ and $$c:=c_1(L)$$. You get the calculation

$$c_t(V\otimes L)=(1+(c(1)+c)t)(1+(c(2)+c)t)=$$

$$...+(c(1)c(2)+(c(1)+c(2))c+c^2)t^2=$$

$$\cdots + (c_2(V)+c_1(V)c_1(L)+\binom{2}{2}c_1(L)^2)t^2=$$

$$c_0(V\otimes L) +c_1(V\otimes L)t+c_2(V\otimes L)t^2.$$

Note: You get

$$c_t(V)=(1+c(1)t)(1+c(2)t)=$$

$$1+(c(1)+c(2))t+c(1)c(2)t^2=c_0(V)+c_1(V)t+c_2(V)t^2$$

hence

$$c_1(V)=c_1(L_1)+c_1(L_2)\text{ and }c_2(V)=c_1(L_1)c_1(L_2).$$

Here you view the elements as "living" in a commutative ring and multiply (they all live in the even cohomology ring $$H^{2*}(X)$$ which is commutative) You find this explained in Hartshorne, Appendix A.

Note: Formulas for Chern classes $$c_i(E\otimes F)$$ where $$E,F$$ have "Chern roots" $$a_i,b_j$$ are expressed in terms of polynomials in the roots $$a_i,b_j$$. The coefficient of the $$t^i$$ term "lives" in the group $$H^{2i}(X)$$. Whenever you have a theory of Chern classes $$c_i(E) \in H^{2i}(X)$$ living in the even part of a cohomology ring $$H^*(X)$$ you may perform such calculations. The experession

$$c_t(E):=c_0(E)+c_1(E)t+\cdots + c_r(E)t^r$$

"lives" in

$$H^0(X)\oplus H^2(X)t\oplus \cdots \oplus H^{2r}(X)t^r.$$

For any rank $$r$$ locally trivial sheaf $$E$$ there is the complete flag bundle $$F(E)$$ of $$E$$ and an injection

$$H^*(X) \subseteq H^*(F(E)),$$

hence you may perform all calculations in $$H^*(F(E))$$. In $$H^*(F(E))$$ you have the equality

$$c_r(E)=c_r(L_1)+ \cdots + c_r(L_r)$$

where $$L_i$$ are invertible sheaves. You get the formula

$$c_t(E)=c_t(L_1)\cdots c_t(L_r)$$

in $$H^{2*}(X)[t]$$: It holds in $$H^{2*}(F(E))[t]$$ but since $$c_t(E) \in H^{2*}(X)[t]$$ the equality holds here.

• Very clear answer. I appreciate it!
– user770533
Dec 3, 2021 at 12:02

In general, there are two ways to compute the Chern classes of a tensor product of complex vector bundles: the splitting principle, and the Chern character. The answer by hm2020 uses the splitting principle, so in this answer I will use the Chern character.

The Chern character satisfies $$\operatorname{ch}(E\otimes F) = \operatorname{ch}(E)\operatorname{ch}(F)$$, and can be expressed in terms of Chern classes as follows

$$\operatorname{ch}(E) = \operatorname{rank}(E) + c_1(E) + \frac{1}{2}(c_1(E)^2 - 2c_2(E)) + \dots$$

where the ellipsis represents terms of higher order (degree six and above) - these terms have no impact on this calculation. So on the one hand we have

$$\operatorname{ch}(V\otimes L) = r + c_1(V\otimes L) + \frac{1}{2}(c_1(V\otimes L)^2 - 2c_2(V\otimes L)) + \dots$$

while on the other, we have

\begin{align*} &\ \operatorname{ch}(V\otimes L)\\ &= \operatorname{ch}(V)\operatorname{ch}(L)\\ &= \left(r + c_1(V) + \frac{1}{2}(c_1(V)^2 - 2c_2(V)) + \dots\right)\left(1 + c_1(L) + \frac{1}{2}c_1(L)^2 + \dots\right)\\ &= r + (c_1(V) + rc_1(L)) + \left(\frac{1}{2}(c_1(V)^2-2c_2(V)) + c_1(V)c_1(L) + \frac{r}{2}c_1(L)^2\right) + \dots \end{align*}

Comparing terms of degree two, we see that $$c_1(V\otimes L) = c_1(V) + rc_1(L)$$ as you've already seen. Comparing terms of degree four, we have

\begin{align*} \frac{1}{2}(c_1(V\otimes L)^2 - 2c_2(V\otimes L)) &= \frac{1}{2}(c_1(V)^2-2c_2(V)) + c_1(V)c_1(L) + \frac{r}{2}c_1(L)^2\\ c_1(V\otimes L)^2 - 2c_2(V\otimes L) &= c_1(V)^2-2c_2(V) + 2c_1(V)c_1(L) + rc_1(L)^2\\ (c_1(V) + rc_1(L))^2 - 2c_2(V\otimes L) &= c_1(V)^2-2c_2(V) + 2c_1(V)c_1(L) + rc_1(L)^2. \end{align*}

Expanding and collecting like terms, we obtain

\begin{align*} -2c_2(V\otimes L) &= -2c_2(V) + (2 - 2r)c_1(V)c_1(L) + (r - r^2)c_1(L)^2\\ c_2(V\otimes L) &= c_2(V) + (r-1)c_1(V)c_1(L) + \frac{r^2-r}{2}c_1(L)^2\\ c_2(V\otimes L) &= c_2(V) + (r-1)c_1(V)c_1(L) + \frac{r(r-1)}{2}c_1(L)^2\\ c_2(V\otimes L) &= c_2(V) + (r-1)c_1(V)c_1(L) + \binom{r}{2}c_1(L)^2.\\ \end{align*}

• Does your proof show that this relation holds at the level of forms (for a given metric)? May 5 at 12:37
• @BinAcker: No it doesn't. Note that this proof works for complex vector bundles over CW complexes; in particular, your question doesn't make sense in this level of generality. If one restricts to smooth complex vector bundles over a smooth manifold, your desired relation may be true, but it would require a different proof. May 5 at 13:21
• Do you know it to be true anyway? May 5 at 13:23
• @BinAcker: Unfortunately I do not know whether it is true or false. May 5 at 13:26