3
$\begingroup$

Let $X$ be a smooth vector field on a manifold $M$. The Jacobian of $X$ at a critical point $x^*$ is the linear map $$X'(x^*): T_{x^*}M \rightarrow T_{x^*}M$$ where $T_{x^*}M$ is the tangent space to $M$ at $x^*$, defined by $$X'(x^*) \cdot u = \frac{d}{dt} \left( \text{d}_{x^*}\Theta_t \cdot u \right)|_{t=0}$$ for any $u \in T_{x^*}M$.

Here $\Theta_t: M \rightarrow M$ is the flow of $X$ and $\text{d}_{x^*}$ is the differential, so $\text{d}_{x^*}\Theta_t$ is a linear map of $T_{x^*}$ to itself, since $x^*$ is a critical point: $$\text{d}_{x^*}\Theta_t: T_{x^*}M \rightarrow T_{\Theta_t(x^*)}M = T_{x^*}M$$

Chosen a local chart at $x^*$, the matrix representation of this linear map is the usual Jacobian matrix: $$\left[ X'(x^*) \right]_{ij} = \left( \frac{\partial X^i}{\partial x^j} \right)_{x = x^*}$$

See e.g. Abraham-Marsden (1978), Foundations of mechanics, p. 72 (page attached below).

Question Is there an analogue coordinates-independent definition for the "Jacobian" of a 1-form $\alpha$ on a manifold $M$ as a linear map on a cotangent space and such that the local representative is the usual Jacobian matrix?

enter image description here

$\endgroup$

1 Answer 1

Reset to default
3
$\begingroup$

The answer is yes, and, in fact, it is a more general notion. The construction is often called the intrinsic derivative. Any time you have a smooth section $s$ of a vector bundle $E\to M$, its derivative at a zero $p$ is a well-defined linear map $Ds_p\colon T_pM\to E_p$.

Of course, you can work in a trivialization on an open set $U$ containing $p$ and then $s$ corresponds to a function $f_U\colon U\to\Bbb R^k$ for $V\subset\Bbb R^n$ and we can consider the derivative $(Df_U)_p\colon T_pM \to \Bbb R^k$. If you have an additional, overlapping trivialization on $V$, then on $U\cap V$ we have $f_V(x)=g_{UV}(x)f_U(x)$, where $g_{UV}\colon U\cap V\to GL(k)$ is the transition function of the vector bundle. In general, $(Df_V)_p = g_{UV}(p)(Df_U)_p + (Dg_{UV})_p f_U(p)$, but at a zero of the section, the second term vanishes and we see that $(Df_U)_p$ defines a well-defined mapping $T_pM\to E_p$.

$\endgroup$
6
  • $\begingroup$ I’ve never heard of this before! The main example I can think of is the Hessian of a function at critical point. But does it appear anywhere else? $\endgroup$ Commented Mar 11, 2023 at 22:41
  • 1
    $\begingroup$ @Deane Just as one computes the index of a vector field at a zero, one can compute this for a section of any oriented vector bundle and then compute the Euler class. I recall encountering it in singularity theory, as well $\endgroup$ Commented Mar 12, 2023 at 0:37
  • $\begingroup$ This is super interesting, thanks @TedShifrin! Any reference on it? How is ${Ds}_{p}$ actually defined? (There's a typo, the section $s$ of a vector bundle is $M \to E$). $\endgroup$ Commented Mar 13, 2023 at 12:11
  • $\begingroup$ There isn’t a typo. $E\to M$ is the vector bundle. I actually learned this in graduate school eons ago and do not know a reference, but I told you how it’s defined :) $\endgroup$ Commented Mar 13, 2023 at 15:47
  • $\begingroup$ P.S. If you prefer, write $T_{(p,0)}E = T_pM\oplus E_p$, and project $Ds(p)$ onto $E_p$. (Only along the zer section do we have a canonical such splitting.) $\endgroup$ Commented Mar 13, 2023 at 16:36

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.