1
$\begingroup$

Technically, a smooth $(k,\ell)$-tensor field over a smooth manifold $M$ should be a member of the space $$\Gamma\left(\coprod_{p\in M} \mathscr T^k_\ell(T_pM)\right)$$ of smooth sections over the $(k,\ell)$-tensor bundle over $M$. However I’ve seen such fields be defined in lectures as multilinear maps $$T: \Gamma(T^*M)^{\times\ell}\times\Gamma(TM)^{\times k} \to C^\infty(M)$$ which I think would have them members of a “space” (module, I guess? but does it make sense for a module to have a dual, and if so would the dual of the module of vector fields really be the module of covector fields?) $$\mathscr T^k_\ell(\Gamma(TM))$$ Why would these two definitions be equivalent? Doesn’t the latter imply that tensor fields have fewer properties than they actually do, since the tensor space is constructed on the module of vector fields, instead of the vector space of tangent vectors?

In other words, does it matter if one goes “tangent space => tensor space => tensor bundle => module of tensor fields” or “tangent space => tangent bundle => module of vector fields => module of tensor fields”?

$\endgroup$
5
  • $\begingroup$ "Multilinear" here means multilinear over $C^{\infty}(M)$. The sheaf of smooth sections of any smooth vector bundle over $M$ is a module over $C^{\infty}(M)$. The corresponding dual module is given by taking $\text{Hom}(-, C^{\infty}(M))$ (Hom here means $C^{\infty}(M)$-linear maps), and you can show that the dual of the sheaf of smooth sections of $V$ is the sheaf of smooth sections of $V^{\ast}$. $\endgroup$ Dec 28, 2017 at 9:54
  • $\begingroup$ Sorry, everywhere I say "sheaf" write "vector space." $\endgroup$ Dec 28, 2017 at 10:06
  • $\begingroup$ Curious about what book you would recommend on the subject for the notation used. $\endgroup$ Dec 28, 2017 at 10:18
  • $\begingroup$ @QiaochuYuan So there is an isomorphism between the two structures? $\endgroup$
    – giobrach
    Dec 28, 2017 at 10:24
  • $\begingroup$ @mathreadler Every time I learn something new and there is notational ambiguity (i.e. always) I choose the one that makes the most sense to me, the one that feels more natural, or the most precise. So, unfortunately, I have no particular books to suggest (if not all of them), although I’m now following Lee’s textbooks on geometry and manifolds. Does my notation make sense to you? $\endgroup$
    – giobrach
    Dec 28, 2017 at 10:30

1 Answer 1

3
$\begingroup$

$\def\lto{\overset{\sim}{\longrightarrow}}$ We have two definitions:

  1. $T \in \Gamma(\bigsqcup_{p\in M} \mathcal T_l^k(T_p M)),$ where $\mathcal T_l^k(V) = \{ \phi: (V^*)^{l \times} \times V^{k \times} \lto F \},$
  2. $T : \Gamma(T^*M)^{l \times} \times \Gamma(TM)^{k \times} \lto C^{\infty}(M, F),$

where $F = \mathbb{R} \text{ or } \mathbb{C}$ and $\lto$ denotes a linear map.

According to definition 1, $T$ practically is a map taking a point $p \in M$ and giving an element in $\mathcal T_l^k(T_p M))$ which itself is a map taking a vectortuple in $(T_p{}^* M)^{l \times} \times (T_p M)^{k \times}$ and returning a number: $$T_{\text{def.1}} : (p \in M) \to (T_p{}^* M)^{l \times} \times (T_p M)^{k \times} \to F.$$

According to definition 2, $T$ instead practically is a map taking first a tuple of functions, $((p \in M) \to T_p^* M)^{l \times} \times ((p \in M) \to T_p M)^{k \times},$ then a point $p \in M$ and returning a number: $$T_{\text{def.2}} : ((p \in M) \to T_p{}^* M)^{l \times} \times ((p \in M) \to T_p M)^{k \times} \to (p \in M) \to F.$$

Thus we have the following relation: $$T_{\text{def.2}}(u_1,\ldots,u_l; u^1,\ldots,u^k)(p) = T_{\text{def.1}}(p)(u_1(p),\ldots,u_l(p); u^1(p),\ldots,u^k(p))$$ and reversely, $$T_{\text{def.1}}(p)(u_1,\ldots,u_l; u^1,\ldots,u^k) = T_{\text{def.2}}(p\mapsto u_1,\ldots,p\mapsto u_l; p\mapsto u^1,\ldots, p\mapsto u^k)(p).$$

(Sorry for some unconventional notation; I hope that it's clear anyway.)

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .