Quick question: Chern classes of Sym, Wedge, Hom, and Tensor
As Asal Beag Dubh says in the comments, the key point is to use the splitting principle to reduce the computations to the case of line bundles. Everything becomes more or less an exercise in symmetric function theory. Here are the four basic examples:
The dual bundle
If $L$ is a line bundle with Chern class $c_1(L)$, then the dual line bundle $L^{\ast}$ is isomorphic to the inverse line bundle and hence has Chern class $c_1(L^{\ast}) = - c_1(L)$. If $V \cong \bigoplus_i L_i$ then $V^{\ast} \cong \bigoplus_i L_i^{\ast}$, so we have
$$c(V^{\ast}) = \prod_i (1 - c_1(L_i))$$
from which it follows that
$$c_i(V^{\ast}) = (-1)^i c_i(V).$$
The tensor bundle
If $L, L'$ are line bundles with Chern classes $c_1(L), c_1(L')$, then the tensor product $L \otimes L'$ has Chern class $c_1(L \otimes L') = c_1(L) + c_1(L')$. If $V \cong \bigoplus_i L_i$ and $V' \cong \bigoplus_j L_j'$, then
$$V \otimes V' \cong \bigoplus_{i, j} L_i \otimes L_j'$$
so we have
$$c(V \otimes V') = \prod_{i, j} (1 + c_1(L_i) + c_1(L_j')).$$
Extracting more explicit formulas from this is a tedious exercise. An alternative here is to use the Chern character, which is multiplicative with respect to tensor product by design:
$$\text{ch}(V \otimes V') = \text{ch}(V)\text{ch}(V').$$
For example, this gives
$$c_1(V \otimes V') = c_1(V) \dim V' + c_1(V') \dim V.$$
You can get corresponding formulas for the hom bundle using the isomorphism $V^{\ast} \otimes W \cong \text{Hom}(V, W)$.
The symmetric powers
It's cleanest to do all of the symmetric powers at once. The key is the isomorphism
$$S(V \oplus W) \cong S(V) \otimes S(W)$$
where $S(V) \cong \bigoplus_i S^i(V)$ is the symmetric algebra. This is an isomorphism of graded vector bundles, and remembering the grading is important in what comes next. If $V \cong \bigoplus_i L_i$, it follows that
$$S(V) \cong \bigotimes_i S(L_i)$$
and hence that the graded Chern character of $S(V)$, as a graded vector bundle, can be computed as
$$\text{ch}(S(V)) = \sum_k t^k \text{ch}(S^k(V)) = \prod_i \text{ch}(S(L_i)) = \prod_i \frac{1}{1 - t e^{c_1(L_i)}}$$
where $t$ is a formal variable. Again, extracting more explicit formulas from this is a tedious exercise.
The exterior powers
As for the symmetric powers, we again have
$$\Lambda(V \oplus W) \cong \Lambda(V) \otimes \Lambda(W)$$
where $\Lambda(V) \cong \bigoplus_i \Lambda^i(V)$ is the exterior algebra. The discussion is exactly the same as for the symmetric algebra except that the last Chern character computation is a bit different, and we get
$$\text{ch}(\Lambda(V)) = \sum_k t^k \text{ch}(\Lambda^k(V)) = \prod_i \text{ch}(\Lambda(L_i)) = \prod_i (1 + t e^{c_1(L_i)}).$$
Once again, extracting more explicit formulas from this is a tedious exercise. To get you started on $c_2(\Lambda^2(V))$, by looking at the coefficient of $t^2$ we get
$$\text{ch}(\Lambda^2(V)) = \sum_{i < j} e^{c_1(L_i) + c_1(L_j)}.$$
The first term of this expansion gives you the dimension of $\Lambda^2(V)$, which you already know. The second term gives you the first Chern class, which is
$$c_1(\Lambda^2(V)) = (\dim V - 1) c_1(V).$$
The third term gives you the third term in the Chern character of $\Lambda^2(V)$, which you need to correct a little by $c_1^2$ to get $c_2$.
See the section "Whitney's Theorem and the Splitting Principle" in the draft of
- Eisenbud and Harris, 3264 and All That: Intersection Theory in Algebraic Geometry.
It's page 60-61 of this PDF:
http://isites.harvard.edu/fs/docs/icb.topic720403.files/book.pdf
In particular, look at Example 1.41.