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!
Best Answer
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<j} (1+x_i+x_j) = 1+x_1+x_2. \end{align*}
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.