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 unoriented 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*}

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*}

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 unoriented 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 $\operatorname{Sq}^1(w_2) = w_3 + w_1w_2$.

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 unoriented 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*}