I am taking courses on functional analysis (through Coursera.com) and differential geometry (textbook author : O'neil) on my university.

I made the following table on my own.

enter image description here

Are the similar concepts linear form and 1-form?

Also I want to know more relationships deeply. Or any recommendation (book or whatever)

(Modified and added) I feel that the weak topology is generated by pulling back in dual space so that we get small open sets. I wonder whether it is the same story as pulling forms back.

  • 5
    $\begingroup$ This is a pretty good observation. They are similar in this way, but there are deeper connections (hah) between the two. Spectral geometry is a really nice synthesis of the two, as is index theory. $\endgroup$ Dec 14, 2014 at 18:18
  • 1
    $\begingroup$ Thank you for remarking Spectral geometry and index theory. $\endgroup$
    – jakeoung
    Dec 14, 2014 at 18:38
  • $\begingroup$ with $E$ supposed a bundle, what do you mean for a linear linear bounded $E\to\Bbb{R}$? $\endgroup$
    – janmarqz
    Dec 15, 2014 at 18:58
  • $\begingroup$ The second diagram would be confusing without description. So I deleted and added a comment. $\endgroup$
    – jakeoung
    Dec 15, 2014 at 19:58
  • $\begingroup$ with a map $F:M\to N$ you induce $F_*:T_pM\to T_{Fp}N$ the derivative and for a $\phi\in T_{Fp}N$ a 1-form you can pullback to get $F_*(\phi)\in T_pM$ $\endgroup$
    – janmarqz
    Dec 16, 2014 at 1:06

1 Answer 1


certainly they are similar concepts. In functional analysis you have a complete normed space $X$ with an inner product $<, >$ i.e. a Hilbert Space. Your tangent space $T_{p}M$ at a point $p\in M$ in a smooth manifold of dimension $n$ is a finite dimensional Hilbert Space $(T_{p}M, <,>_{g})$ with the inner product $<,>_{g}$ induced by the Riemannian metric $g$ hence all the concepts you learned in functional analysis apply to the tangent space at each point.

As you mentioned the dual space is simple the space the of continuous linear forms on $X$ and the concepts translate immediately to the tangent space $T_{p}M$ creating the dual space i.e. the cotangent space $T^{\ast}_{p}M$.

Now in differential geometry we learned the importance of globally defined, invariant objects. Let's defined global Hilbert spaces. Consider the space of (measurable w.r.t. $dV_{g}$) vector fields $\mathcal{F}(M, TM)$. For simplicity let's assume that $M$ is compact. The standard way to define an inner product on $\mathcal{F}^{}(M, TM)$ is by using the integral defined by using the volume form $dV_{g}$ induced by the metric $g$. i.e. $$<X,Y>_{g}=\int_{M}<X_{p},Y_{p}>dv_{g}.$$

It is easy to prove that the closure of $(\mathcal{F}^{}(M, TM), \parallel\cdot\parallel_{g})$ is a Hilbert Space usually denoted $$L^{2}(M,TM)$$ i.e. the classical $L^2$ Hilbert spaces in the context of differential geometry. Of course it is possible to define $L^2$ spaces on any tensor bundle $TM^{\otimes k}\otimes T^{\ast}M^{\otimes s}$. Moreover, if the manifold is not compact, then there are many ways to define Hilbert spaces too by imposing regularity conditions. This is very useful to apply functional anaylsis methods to study PDE on manifolds. A nice introduction to these topics can be found in Chapter 10 "Elliptic Equations on Manifolds" of the book Geometry of Manifolds by L. Nicolaescu.

  • $\begingroup$ Compactness of $M$ is not used in your definition. (You assume that the vector fields/sections are $L^2$-integrable anyway. In the noncompact case just use compactly supported sections. $\endgroup$
    – user99914
    Oct 31, 2017 at 3:22
  • $\begingroup$ yeah, I just used compactness to defined the $L^2$ space directly and avoid defining $L^2$ as a subspace of $\mathcal{F}$ for the sake of clarity in the exposition. In the non-compact case the most straight forward definition would be done by using compactly supported fields as you mentioned but I just wanted to mentioned that there are more useful definitions using the notion of regularity. $\endgroup$
    – Coffee
    Dec 8, 2017 at 2:33

You must log in to answer this question.

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