Does $T\mathbb{C}P^n$ have one-dimensional subbundles?

Question

Does $T\mathbb{C}P^n$ have one-dimensional subbundles when $n>1$? It seems like Chern classes doesn't give any obstacles for this, because of $c(\mathbb{C}P^n)=(1+a)^{n+1}$ and we can conclude only that if such bundle exists then it have to be isomorphic to tautological line bundle of $\mathbb{C}P^n$ (because of $(1+a)$ divides $(1+a)^{n+1}$).

Answer 1

If $E$ is a smooth complex vector bundle and $F$ is a smooth complex subbundle, then we can equip $E$ with a hermitian metric to construct $F^{\perp}$ and obtain the isomorphism $E \cong F \oplus F^{\perp}$.

Every line bundle on $\mathbb{CP}^n$ is isomorphic to $\mathcal{O}(k)$ which has first Chern class $c_1(\mathcal{O}(k)) = ka$, so if $T\mathbb{CP}^n$ has a line subbundle, we see that $T\mathbb{CP}^n \cong \mathcal{O}(k)\oplus E$ where $E$ is a complex vector bundle of rank $n - 1$.

In particular, if $T\mathbb{CP}^2$ has a line subbundle $\mathcal{O}(k)$, there is another line bundle $\mathcal{O}(l)$ such that $T\mathbb{CP}^2 \cong \mathcal{O}(k)\oplus\mathcal{O}(l)$. Note that

$$3a = c_1(T\mathbb{CP}^2) = c_1(\mathcal{O}(k)\oplus\mathcal{O}(l)) = c_1(\mathcal{O}(k)) + c_1(\mathcal{O}(l)) = ka + la = (k + l)a$$

and

$$3a^2 = c_2(T\mathbb{CP}^2) = c_2(\mathcal{O}(k)\oplus\mathcal{O}(l)) = c_1(\mathcal{O}(k))c_1(\mathcal{O}(l)) = (ka)(la) = kla^2$$

so $k + l = 3$ and $kl = 3$. But there are no such integers $k$ and $l$, so $T\mathbb{CP}^2$ does not admit a line subbundle.

As for the general case, we have the following theorem from Glover, Homer, & Stong - Splitting the Tangent Bundle of Projective Space (Theorem $1$ (ii)).

If $n$ is even, $T\mathbb{CP}^n$ does not split into a non-trivial Whitney sum of complex subbundles.

If $n$ is odd, $T\mathbb{CP}^n$ splits only into a complex line bundle and its complement.