4

Let $M$ be a smooth manifold and $\pi \colon TM \to M$ denote its tangent bundle. Considering a tangent space $TTM$ to its tangent space $TM$ at a point $(p,u) \in TM$, I would intuitively expect that $$T_u TM = T_u T_p M = T_{(p,u)} TM$$ are the same thing since $p = \pi(u)$ anyway. But the dimensions do not add up since $TM$ is 2-dimensional and $T_u TM$ therefore as well, while $T_p M$ is n-dimensional and $T_u T_p M$ must therefore be as well n-dimensional.

Since $TTM$ is difficult to think about or to visualize, I cannot see the difference between two vector spaces.

qarabala
  • 1,131

1 Answers1

6

$T_u(T_pM)$ is called the vertical space of $TM$ at $u$. It is only part of the tangent space $T_u(TM)$.

More generally, if you have a fiber bundle $(X,\pi,B)$ and a point $b\in B$ then we can consider the fiber $X_b:=\pi^{-1}(\{b\})$ (this is a submanifold of $X$ of dimension $\dim X - \dim B$, since $\pi$ is a submersion). Now, at a point $x\in X_b$ in the fiber, we can consider two vector spaces:

  • $T_xX$; the tangent space to $X$ at $x$
  • $T_x(X_b)$; the tangent space to the fiber $X_b$ at $x$. One can easily check (since $X_b$ is the $b$-level set of $\pi$) that this equals $\ker (T\pi_x)$. One often denotes this vector space as $V_xX$, called the vertical space of $(X,\pi,B)$ at $x$.

The latter is a subspace of the former, and it should make sense intuitively why that is: when we take tangent vectors to a submanifold, we’re only looking at velocity vectors of curves contained in that submanifold, i.e there are fewer directions to go in. And as for why we use the term “vertical” it’s because we like to imagine fiber bundles as a bunch of straws arranged vertically (each straw representing the fiber). Nothing deep here.

So even intuitively there’s no reasom why you should expect them to be the same thing. The only thing which I am mildly sympathetic to is if you wanted to say something like $T_v(T_wM)\cong T_w(T_vM)$ (a-priori it’s nonsense but given that one has symmetry of partial derivatives from multivariable calculus, this is a natural sort of thing to think about). Of course this doesn’t even make sense “type-wise”, but in the special case that $M$ is a vector space itself, both these spaces csn be canonically identified with $M$ itself.

peek-a-boo
  • 65,833