1
$\begingroup$

In the case of a finite-dimensional vector space $V$ over some arbitrary field $\mathbb{F}$ it is well-known that there is the following canonical isomorphism:

$$\underbrace{V\otimes_{\mathbb{F}} \dots \otimes_{\mathbb{F}} V}_{r-\text{times}}\otimes_{\mathbb{F}} \underbrace{V^{\ast}\otimes_{\mathbb{F}}\dots\otimes_{\mathbb{F}} V^{\ast}}_{s-\text{times}}\overset{\text{can.}}{\cong} L^{r+s}(\underbrace{V^{\ast},\dots,V^{\ast}}_{r-\text{times}},\underbrace{V,\dots,V}_{s-\text{times}},\mathbb{F})$$

where $L^{r+s}(V^{\ast},\dots,V^{\ast},V,\dots,V,\mathbb{F})$ denotes the set of all multilinear maps from $(V^{\ast})^{r}\times V^{s}$ to $\mathbb{F}$.

Is the same also true for a module $M$ over some ring $R$? And if the answer is yes, does we need to assume that the ring is commutative, or does it hold in general?

(I ask because i would like to unterstand the equality between the set of all tensor fields and the set $$\underbrace{\mathfrak{X}(\mathcal{M})\otimes_{C^{\infty}(\mathcal{M})} \dots \otimes_{C^{\infty}(\mathcal{M})} \mathfrak{X}(\mathcal{M})}_{r-\text{times}}\otimes \underbrace{\mathfrak{X}^{\ast}(\mathcal{M})\otimes_{C^{\infty}(\mathcal{M})}\dots\otimes_{C^{\infty}(\mathcal{M})} \mathfrak{X}^{\ast}(\mathcal{M})}_{s-\text{times}}$$ for some smooth manifold $\mathcal{M}$.)

$\endgroup$
4
  • 1
    $\begingroup$ For a module $M$ over a general ring $R$ what analogue of "finite dimensional" do you want? Finitely generated? Finitely presented? ACC on submodules? DCC on submodules? Also, you need to restrict to the commutative case since your formulas aren't even interpretable when $R$ is non-commutative (unless you want $R$ to be a non-commutative algebra over a commutative ring $k$ over which the tensor products are taken or unless you want $M$ to be a bimodule). $\endgroup$
    – C Monsour
    Mar 20, 2021 at 20:36
  • 1
    $\begingroup$ What does finite-dimensional mean over a ring? $\endgroup$
    – J. De Ro
    Mar 20, 2021 at 20:37
  • $\begingroup$ Okay I see.....I don't have too much background in modules..... $\endgroup$
    – B.Hueber
    Mar 20, 2021 at 20:39
  • $\begingroup$ But then I don't unterstand the relation of tensor fields to the set written above. Because what I know is that a tensor field (defined as a section of the tensor bundle) can be identified with a $C^{\infty}$-multilinear map with values in some copy of the set of vector fields and covectorfield. How is this then related to the tensor product written above? $\endgroup$
    – B.Hueber
    Mar 20, 2021 at 20:41

1 Answer 1

Reset to default
3
$\begingroup$

You need the ring to be commutative if you want to consider tensor products of modules, otherwise you can only define the tensor product of a right $R$-module by a left $R$-module and what results cannot generally be endowed with a structure of $R$-module.

So, let's assume $R$ is a commutative ring, so for any two modules $M,N$, $M\otimes_RN$ can be given a structure of $R$-module.

For a starter, you wish that $V\cong L(V^*,R)$, where $V^*=L(V,R)$. Unfortunately this holds quite rarely, even for finitely generated modules over a Noetherian ring. For instance, if $V$ is a finitely generated torsion group (a module over $\mathbb{Z}$), then $L(V,\mathbb{Z})=0$.

You have to restrict to finitely projective modules, in order that the thing works.

$\endgroup$
2
  • $\begingroup$ Thanks a lot for the answer. So for finitely generated projective modules this holds in general? To come back to my original problem on Tensor fields: Does it apply in this case, i.e. can we identify $\mathfrak{X}(\mathcal{M})^{\otimes_{C^{\infty}(\mathcal{M})}r}\otimes_{C^{\infty}(\mathcal{M})}\mathfrak{X}^{\ast}(\mathcal{M})^{\otimes_{C^{\infty}(\mathcal{M})}s}$ with the set of all $C^{\infty}(\mathcal{M})$-multilinear maps. I guess yes, because a quick google search results in "Serre–Swan theorem", which states that the set of all vector fields is a finitely-generated projective module..... $\endgroup$
    – B.Hueber
    Mar 20, 2021 at 21:24
  • 1
    $\begingroup$ @Udalricus.S. That nails it! :-) $\endgroup$
    – egreg
    Mar 20, 2021 at 21:39

You must log in to answer this question.

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