# About Chern classes of symmetric powers

I have the following question

I'm training in computation of Chern classes. Let $$\xi$$, $$\eta$$ are complex vector bundles of rank two ($$r=2$$)

I'm trying to find $$c_1(S^2 \xi \otimes \eta)$$, $$c_2(S^2 \xi \otimes \eta)$$, $$c_3(S^2 \xi \otimes \eta)$$ where $$S$$ is a symmetric power in terms of Chern classes of $$\eta$$ and $$\xi$$.

My attempt is the following: in general case $$c_1(E\otimes F) = rc_1(F) + sc_1(E)$$ where $$r$$ and $$s$$ are ranks of vector bundles. In our case we have $$c_1(S^2 \xi \otimes \eta)=2c_1(S^2 \xi)+2c_1(\eta)$$ and I don't know how to define $$c_1(S^2 \xi)$$ in terms of chern classes of $$\xi$$ vector bundle.

For the second chern class in general case for vector bundles of ranks $$s$$ and $$n$$ we have the following equation

$$c_2(E\otimes F) = rc_2(F) + sc_2(E) + \binom{r}{2}c_1(F)^2 + \binom{s}{2}c_1(E)^2 + (rs-1)c_1(E)c_1(F)$$.

In our case we have obtained that

$$c_2(S^2 \xi \otimes \eta)=2c_2(S^2 \xi)+2c_2(\eta)+c_1(\eta)^2+c_1(S^2 \xi)^2+3c_1(S^2 \xi)c_1(\eta)$$

For the third Chern class we have that

$$c_3(E\otimes F)=3\textstyle\binom{s}{3}c_1(E)^3+ 3\binom{r}{3}c_1(F)^3 + 6\binom{s}{2}c_1(E)c_2(E)+ 6\binom{r}{2}c_1(F)c_2(F)\\ + 3sc_3(E) + 3rc_3(F)+3(rs-2)c_2(E)c_1(F)+3(rs-2)c_1(E)c_2(F)\\ +\left(\tfrac{3}{2}rs - 1\right)(s-1)c_1(E)^2c_1(F) + \left(\tfrac{3}{2}rs-1\right)(r-1)c_1(E)c_1(F)^2.$$

In our case for ranks $$r=s=2$$ we have obtained that

$$c_3(S^2 \xi \otimes \eta)=\textstyle c_1(S^2 \xi)^3+ c_1(\eta)^3\\ +6c_1(S^2 \xi)c_2(S^2 \xi) +6c_1(\eta)c_2(\eta) + 6c_3(S^2 \xi) +6c_3(\eta)+6c_2(S^2 \xi)c_1(\eta)+6c_1(S^2 \xi)c_2(\eta) +5c_1(S^2 \xi)^2c_1(\eta) + 5c_1(S^2 \xi)c_1(\eta)^2.$$

I hope that i'm right in these computations. I don't know how to define $$c_1(S^2\xi)$$, $$c_2(S^2\xi)$$, $$c_3(S^2\xi)$$ in terms of Chern classes of $$\xi$$ bundle. Can you help me please and explain it in more details if you don't mind. I just know from Bott Tu book that $$c(S^{p} E)=\prod_{1 \leq i_1 \leq i_2 \leq \ldots \leq i_p \leq n} (1+x_{i_1}+\ldots+x_{i_p})$$ and I don't know how to apply this formula for computation these classes. Also I want to know what happened in the case $$c_1(\Lambda^2\xi)$$, $$c_2(\Lambda^2\xi)$$, $$c_3(\Lambda^2\xi)$$ where $$\Lambda$$ is an exterior power

Thank you!

• Use the splitting principle everywhere. May 28 at 14:07
• @Sasha I understand this. Can you show please if you don't mind in example of $c_3(S^2\xi)$? Because I can't find any examples of computations. Thank you. May 28 at 14:12
• Look, for example, at Porteous's Simple Singularities of Maps or Harris and Tu's "On symmetric and skew-symmetric determinantal varieties" (Topology (23) 1984). May 28 at 17:23
• @TedShifrin Thank you for these references. It doesn't help me but it's interesting. May 28 at 19:25
• Sorry. I thought that Harris/Tu in particular derived such formulas in their paper. I think you can also figure it out from Hirzebruch's Topological Methods in Algebraic Geometry. May 28 at 19:34

For small rank bundles, it's not hard to write this out explicitly. As Bott-Tu suggested (and as Sasha reiterated), you use the splitting principle and formally write $$\xi = L_1\oplus L_2\oplus\dots\oplus L_k$$ with $$c_1(L_j) = x_j$$. Then, with $$k=2$$, we have \begin{align*} c(\xi\otimes\xi) &= \prod_{i,j} (1+x_i+x_j) = (1+2x_1)(1+x_1+x_2)(1+x_2+x_1)(1+2x_2);\\ c(S^2\xi) &= \prod_{i\le j} (1+x_i+x_j) = (1+2x_1)(1+x_1+x_2)(1+2x_2);\\ c(\Lambda^2\xi) &= \prod_{i

As a double-check for what follows, you know that $$0\to S^2(\xi) \to \xi\otimes\xi \to \Lambda^2(\xi)\to 0$$ is a short exact sequence, so $$c(\xi\otimes\xi) = c(S^2\xi)c_(\Lambda^2\xi)$$ must hold.

We read off $$c_\ell$$ of a bundle by taking the monomials of total degree $$\ell$$ in its Chern polynomial. In particular, \begin{align*}{} c_1(\xi\otimes\xi) &= 4(x_1+x_2)=4c_1(\xi),\\ c_1(S^2\xi) &= 3(x_1+x_2) = 3c_1(\xi) \\ c_1(\Lambda^2\xi) &= x_1+x_2 = c_1(\xi). \end{align*} In particular, $$c_1(\xi\otimes\xi) = 4c_1(\xi) = c_1(S^2\xi)+c_1(\Lambda^2\xi)$$, as required.

Next, \begin{align*} c_2(\xi\otimes\xi) &= (2x_1)(2x_2)+5(x_1+x_2)^2 = 5c_1^2(\xi)+4c_2(\xi) \\ c_2(S^2\xi) &=2(x_1+x_2)^2+4x_1x_2=2c_1^2(\xi)+4c_2(\xi) \\ c_2(\Lambda^2\xi) &= 0 \end{align*} As a check, $$c_2(\xi\otimes\xi) = c_2(S^2\xi) + c_2(\Lambda^2\xi) + c_1(S^2\xi)c_1(\Lambda^2\xi)$$.

Finally, \begin{align*} c_3(\xi\otimes\xi) &= 2(x_1+x_2)^3+8(x_1+x_2)x_1x_2 = 2c_1^3(\xi)+8c_1(\xi)c_2(\xi)\\ c_3(S^2\xi) &= 4(x_1+x_2)x_1x_2 = 4c_1(\xi)c_2(\xi) \\ c_3(\Lambda^2\xi) &= 0 \end{align*} As a last check, $$c_3(\xi\otimes\xi) = c_3(S^2\xi) + c_2(S^2\xi)c_1(\Lambda^2\xi)$$, as it should.

• It's a perfect answer. Thank you very much and thank you for your time. Also I want to ask you about my first part of computation (where I have defined $c_1(S^2 \xi \otimes \eta)$, $c_2(S^2 \xi \otimes \eta)$, $c_3(S^2 \xi \otimes \eta)$ because I'm not sure about them. Are these equations correct or not? May 28 at 22:52
• No, your formulas are off. If $\xi$ and $\eta$ are both rank $2$, then $E=S^2\xi$ has rank $3$ and $F=\eta$ has rank $2$. So you need to adjust your computations accordingly. Again, the easiest way to do all this is to write out $c(E\otimes F) = \prod_{1\le i\le 3, 1\le j\le 2} (1+x_i+y_j)$ and collect terms. May 28 at 23:00
• Thank you. Are my mistakes in coefficients as i understand? And this sequences $0\to S^p(\xi) \to \xi\otimes\xi\otimes \xi\ldots \to \Lambda^p(\xi)\to 0$ will be exact for p symmetric power? And as a result $c(\xi\otimes\xi\otimes \xi \ldots) = c(S^p\xi)c_(\Lambda^p\xi)$ for p degrees or not in general case? May 28 at 23:08
• No, the symmetric and skew-symmetric parts make up the whole tensor product only in the case of the square. For higher order tensor products, there are all sorts of mixed symmetries. In this case, the stuff we did with the splitting principle works fine; my short exact sequence check cannot be applied, however. May 28 at 23:10
• @Victory: You can use the splitting principle to work out $c_i(S^p\xi)$ in general. See here for $i = 1, 2$ (this is the case where $\operatorname{rank}\xi = 2$, but the same method can be used for any rank). May 31 at 14:12