In general it is not true that a vector bundle $E$ is isomorphic to its dual bundle $E^*$. But it is true when the vector bundle is the tangnet space of a manifold (at least I think it's true). How does one prove this?

# [Math] Tangent Bundle and its (Isomorphic?) Dual Bundle

differential-geometry

#### Related Solutions

$\mathbf P^1(\mathbb R)$ is a closed curve, so its tangent bundle is trivial. There is not much to be imagined, really, in that case! In general, what kind of answer do you expect for your «what is the tangent bundle of $\mathbf P^n$?». It is the tangent bundle of $\mathbf P^n$...

Regarding your last question: if a group $G$ acts properly discontinuously and differentiably on a manifold $M$, then $M/G$ is a manifold, $G$ acts on $TM$ in a natural way (also properly discontinuously, and morover linearly on fibers), and $(TM)/G$ is "the same thing" as $T(M/G)$.

I think the best answer for your general question comes from observing that $\pi:X\to X/G$ is a surjective submersion. Hence for each $x\in X$, you can identify the tangent space $T_{\pi(x)}X/G$ with the quotient of $T_xX$ by the tangent space of the orbit through $x$. The latter is the subspace of $T_xX$ spanned by the fundamental vector fields for the action associated to elements in the Lie algebra $\mathfrak g$ of $G$. Here for $A\in\mathfrak g$, the fundamental vecor field is defined by $\zeta_A(x)=\frac{d}{dt}|_{t=0}x\cdot exp(tA)$. For $\mathbb CP^n$ realized as a quotient of $S^{2n+1}$, this gives you and identification of the tangent space to the complex line spanned by $x$ as the quotient of $x^\perp$ (real orthocomplement) by imaginary multiples of $x$. This is the right space, but not quite the identification that you would like to get.

To get an identification like the one you want, you probably have to realize $\mathbb CP^n$ as a homogeneous space (since this allows you to compare tangent spaces at different points to some extent). The most general version of this is viewing it as a homogeneous space of $GL(n,\mathbb C)$. Then for each line $D\subset\Bbb C^{n+1}$, the map $A\mapsto A(D)$ defines a surjective submersion from $GL(n,\Bbb C)$ onto $\mathbb CP^n$. Hence you get an identification of $T_D\mathbb CP^n$ with the quotient of the tangent space of $GL(n,\mathbb C)$ (which is just the space of complex $n\times n$-matrices) by the Lie algebra of the stabilizer of $D$, which is simply formed by all matrices mapping $D$ to itself. This quotient can be identified with the space of linear maps from $D$ to $\mathbb C^{n+1}/D$, and this is the identification you want. (If you prefer to involve a complex orthocomplement, you can set up a similar picture with $U(n)$ or $SU(n)$ acting on the space of lines.)

## Best Answer

An isomorphism between a vector bundle and its dual, fibre-by-fibre is just an isomorphism between a fibre and its dual. An isomorphism between a vector space and its dual is given by a non-degenerate bilinear function -- so for example, an inner product suffices. So if you had an inner product on your vector bundle, you would have an isomorphism between $E$ and $E^*$ simply by the operation $v \longmapsto \langle v, \cdot \rangle$.

Generally speaking, vector bundles have inner products. For example, if the base space is paracompact and the fibres are finite-dimensional.