2

Let $f: M \longrightarrow M'$ be a diffeomorphism between two smooth manifolds. We denote by $Vect(M)$ the space of smooth vector fields (ie. derivations) on $M$. Then, the map $f_*: Vect(M) \longrightarrow Vect(M')$ defined by $(f_*X)_{m'}=d_{f^{-1}(m')}fX_{f^{-1}(m')}$ is a Lie algebra homomorphism, that is $f_*([X,Y])=[f_*(X), f_*(Y)]$.

Furthermore, a diagram is given, where $TM$ denotes the tangent bundle to $M$. $f_*X$ is said to be the only vector field making it commute.

$$\require{AMScd} \begin{CD} M @>{f}>> M'\\ @VXVV @VVf_*XV \\ TM @>{df}>> TM' \end{CD}$$

I don't get the definition of $f_*$. So, on the left side, we have a vector field $X$ in $Vect(M)$, to which we associate another vector field $f_*X$, by applying a homomorphism, and finally we apply it at some point $m'$ in $M'$. But the right side is not quite clear to me. With the diagram, I see how it is equivalent, but I don't get what it intuitively does.

Spida
  • 359
  • 1
  • 10
  • 2
    The differential of the diffeomorphism $f$ maps vector fields on $M$ to vector fields on $M'$. However, the "naive way" of writing the vector field would be $M \ni p \mapsto (df)p (X_p)$. This, however, is not a vector field on $M'$, as it takes as input points in $M$. To remedy this (this is where the assumption that $f$ is a diffeomorphism is important), we do it "backwards". Take a point $q \in M'$. This has a well-defined unique preimage, $f^{-1}(q)$. We then consider the vector field $M' \ni q \mapsto (df){f^{-1}(q)} (X_{f^{-1}(q)})$, and now this is indeed a vector field on $M'$. – C_M Oct 27 '21 at 14:13
  • Understood! Thanks a lot! – Spida Oct 27 '21 at 14:21

0 Answers0