References for Stiefel-Whitney class of Stiefel manifolds and Grassmannians

The Stiefel manifolds are all parallelizable for $n-k\ge2$, so their total Stiefel-Whitney classes are equal to $1$. A reference is Theorem 3.1 of this paper by Kee Yuen Lam.

For the finite Grassmannians, things are a little more complicated. In the complex and symplectic cases you should be able to calculate these using the Wu formula. Partial results on the normal Stiefel-Whitney classes $\bar{w}(G_{n-k}(\mathbb{R}^n))$ are scattered throughout the literature, a recent reference being

Korbaš, J.; Novotny, P. On the dual Stiefel-Whitney classes of some Grassmann manifolds. Acta Math. Hungar. 123 (2009), no. 4, 319–330.

This article and the articles it references are content with showing non-triviality of some normal SW-class in order to deduce non-immersion results, and do not give formulae for $w(G_{n-k}(\mathbb{R}^n))$.

Most of the methods seem to use the vector bundle isomorphisms \begin{align*} T(G_{n-k}(\mathbb{R}^n)) & \cong \operatorname{Hom}(\gamma_k,\gamma_{n-k})\\ & \cong \gamma_k^*\otimes \gamma_{n-k} \\ & \cong \gamma_k\otimes \gamma_{n-k}\end{align*} and the splitting principle.

In this note, the first two Stiefel-Whitney classes of unoriented, oriented, and complex grassmannians are determined in terms of the Stiefel-Whitney classes of their tautological bundles. This is achieved via the method mentioned at the end of Mark Grant's answer.

For the unoriented grassmannian $\operatorname{Gr}(m, m+n) = O(m+n)/(O(m)\times O(n))$, we have

\begin{align*} w_1(\operatorname{Gr}(m, m+n)) &= (m + n)w_1(\gamma)\\ w_2(\operatorname{Gr}(m, m + n)) &= \left[\binom{m}{2} + \binom{n}{2} + m^2 + mn - 1\right]w_1(\gamma)^2 + (m^2 + n^2)w_2(\gamma)\\ &= \begin{cases} 0 & m - n \equiv 2 \bmod 4\\ w_2(\gamma) & m - n \equiv 1 \bmod 4\\ w_1(\gamma)^2 & m - n \equiv 0 \bmod 4\\ w_2(\gamma) + w_1(\gamma)^2 & m - n \equiv 3 \bmod 4. \end{cases} \end{align*}

For the oriented grassmannian $\operatorname{Gr}^+(m, m+n) = SO(m+n)/(SO(m)\times SO(n))$, we have

\begin{align*} w_1(\operatorname{Gr}^+(m, m + n)) &= 0\\ w_2(\operatorname{Gr}^+(m, m + n)) &= \begin{cases} 0 & m - n \equiv 0 \bmod 2\\ w_2(\gamma_+) & m - n \equiv 1 \bmod 2. \end{cases} \end{align*}

For the complex grassmannian $\operatorname{Gr}^{\mathbb{C}}(m, m + n) = U(m + n)/(U(m)\times U(n))$, we have

\begin{align*} w_1(\operatorname{Gr}^{\mathbb{C}}(m, m + n)) &= 0\\ w_2(\operatorname{Gr}^{\mathbb{C}}(m, m + n)) &= (m + n)w_2(\gamma_{\mathbb{C}}). \end{align*}

As for the quaternionic grassmanian $\operatorname{Gr}^{\mathbb{H}}(m, m + n) = Sp(m + n)/(Sp(m)\times Sp(n))$, it is $2$-connected, so $w_1(\operatorname{Gr}^{\mathbb{H}}(m, m + n)) = 0$ and $w_2(\operatorname{Gr}^{\mathbb{H}}(m, m + n)) = 0$.

In principle, you can calculate all of the Stiefel-Whitney classes using the splitting principle argument, but the calculations get more and more complicated. A short cut for the third Stiefel-Whitney class is to use the fact that $\operatorname{Sq}^1(w_2) = w_3 + w_1w_2$. It follows that we have

\begin{align*} w_3(\operatorname{Gr}(m, m + n)) &= \begin{cases} 0 & m - n \equiv 0, 2 \bmod 4\\ w_3(\gamma) & m - n \equiv 1 \bmod 4\\ w_3(\gamma) + w_1(\gamma)^3 & m - n \equiv 3 \bmod 4 \end{cases}\\ & \\ w_3(\operatorname{Gr}^+(m, m + n)) &= \begin{cases} 0 & m - n \equiv 0 \bmod 2\\ w_3(\gamma_+) & m - n \equiv 1 \bmod 2 \end{cases}\\ & \\ w_3(\operatorname{Gr}^{\mathbb{C}}) &= 0\\ & \\ w_3(\operatorname{Gr}^{\mathbb{H}}) &= 0. \end{align*}