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!

  • $\begingroup$ 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 $\endgroup$
    – user770533
    Dec 3, 2021 at 11:53

2 Answers 2


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

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


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

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

Note: You get

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



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

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.

  • $\begingroup$ Very clear answer. I appreciate it! $\endgroup$
    – 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*}

  • $\begingroup$ Does your proof show that this relation holds at the level of forms (for a given metric)? $\endgroup$
    – BinAcker
    May 5 at 12:37
  • $\begingroup$ @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. $\endgroup$ May 5 at 13:21
  • $\begingroup$ Do you know it to be true anyway? $\endgroup$
    – BinAcker
    May 5 at 13:23
  • 1
    $\begingroup$ @BinAcker: Unfortunately I do not know whether it is true or false. $\endgroup$ May 5 at 13:26

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