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There are several ways we can define tangent space at point $p$ on a manifold $M$.

Definition-1: $$T^{\text{glue}}M=\sqcup_i(U_i\times\mathbb R^n)/\sim$$ where for $(x,v)\in U_i\times\mathbb R^n$, $(y,w)\in U_j\times\mathbb R^n$ we have $(x,v)\sim(y,w)$ if and only if $y=\phi_j\phi_i^{-1}(x)$ and $w=d(\phi_j\phi_i^{-1})_x(v)$. We define the tangent space at $p\in M$ as $T^{\text{glue}}_pM=\{[p,v]:v\in\mathbb R^n\}$ and $\pi:T^{\text{glue}}M\rightarrow M,\pi([p,v])=p$.

Definition-2: $$T^{\text{path}}_pM=\{\text{paths }\gamma:(-\epsilon,\epsilon)\rightarrow M:\gamma(0)=p\}/\sim$$ where $\alpha\sim\beta$ if $(\phi_i\circ\alpha)'(0)=(\phi_i\circ\beta)'(0)$ for every $i$ such that $p\in U_i$. $T^{\text{path}}M=\sqcup_{p\in M} T^{\text{path}}_pM$ and $\pi:T^{\text{path}}M\rightarrow M,\pi(\gamma)=\gamma(0)$.

Definition-3: A derivation of germs is an $\mathbb R$-linear map $X:\operatorname{Germs}_p\rightarrow \mathbb R$ which satisfies, $$X(fg)=f(p)X(g)+X(f)g(p)$$ We define $T^{\text{der}}_p M$ to be the set of derivations of germs at $p$.

Maybe I can show the equivalence of Definition 1 and 2 using,

$$ \begin{align*} \Phi:T^{\text{path}}_p M &\rightarrow T^{\text{glue}}_p M\\ [\gamma]&\mapsto [\phi_i\circ\gamma(0),(\phi_i\circ\gamma)'(0)] \end{align*} $$

Question 1: How to show the equivalence for derivation definition with other ones?


So far, I understand all the definitions of the tangent space. But I was wondering how to define the cotangent space in the context of each definition. Most the resource which I follow define the cotangent space as the dual vector space of the tangent space, like in the Differential Geometry by Loring W.Tu, differential 1-from was defined as,

A $1$-form on $M$ is the assignment of a linear function $\omega_p:T_p M\rightarrow \mathbb R$ to each $p$ in $M$.

I got another weird definition of $1$-form (Though it was for holomorphic $(1,0)$-form, source: 4.3.2, Riemann surfaces by Dror Varolin, I just replace holomorphic term with differentiable).

If $f_i:U_i\rightarrow\mathbb R$ are differentiable functions on domain $U_i$ containing $p$, we say that $(f_1,U_1)$ and $(f_2,U_2)$ define the same germ if there is an open subset $U\subset U_1\cap U_2$ such that $f_1\mid_U=f_2\mid_U$. The equivalence class of all $(f,U)$ that define the same germs is denoted by $[f]_p$ and is called a germ of a differentiable function. The set of all germs of differential functions is denoted by $\mathcal O_{M,p}$. I (guess I) knew that it is the sheaf of differentiable function at point $p$. Now they define the $1$-form as,

Two germs $[f]_p=(f,U)$ and $[g]_p=(g,V)$ are said to be cotangent, written $[f]_p\sim_1 [g]_p$, if there is a coordinate chart $\varphi:W\rightarrow\mathbb R$ such that $p\in W\subset U\cap V$ and $$(f\circ\varphi^{-1})'(\varphi(p))=(g\circ\varphi^{-1})'(\varphi(p))$$
The set of equivalence classes of mutually tangent germs is denoted by $K_{M,p}$, and the elements of $K_{M,p}$ are called $1$-form. The equivalence class of germs tangent to $[f]_p$ is denoted by $df(p)$.

Question 2: I guess I couldn't relate to this definition or failed to understand its motivation. How is this definition related to the covector?

I was searching for the definition of cotangent space in the context of Definition-1,2 and 3 given for tangent space above. Because I want to know how the dualization idea arises for each case, another reason is, that we can show the equivalency of each definition for tangent space, I want to see the same thing for cotangent space $(T^{\text{glue}}M)^*$, $(T^{\text{paths}}M)^*$ and $(T^{\text{der}}M)^*$.

Question 3: How do we define the duality for each definition mentioned for tangent space above?

Any resource/help will be appreciated as I was struggling with these too much. Though I restricted everything to smooth manifold settings, if anyone enlightens me about complex manifolds, you are welcome.


I understand my question 2, hence remove it. Hence, question-3 is the only one which I want to resolve.

For question-3, I guess couldn't come up with natural ideas in the following places:

  • $T^{\text{glue}*}_pM=\{\omega_p:T^{\text{glue}}_pM\rightarrow\mathbb R\}$, then for any $[p,v]\in T^{\text{glue}}_pM$, $\omega_p([p,v])=?$
  • $T^{\text{path}*}_pM=\{\omega_p:T^{\text{path}}_pM\rightarrow\mathbb R\}$, then for any $[\gamma]\in T^{\text{path}}_pM$, $\omega_p([\gamma])=?$
  • $T^{\text{der}*}_pM=\{\omega_p:T^{\text{der}}_pM\rightarrow\mathbb R\}$, then for any $X\in T^{\text{der}*}_pM$, $\omega_p(X)=?$

Okay, after getting the dual vector space how to glue them together to construct the cotangent bundle? As the definition 1,2,3 we have some sort of equivalence relation which helps us for gluing. But now, I don't see something naturally arising from these information.

  • With the definition using derivations, the cotangent space $T_p^*M$ is naturally isomorphic to ${\mathfrak m}/{\mathfrak m}^2$, where ${\mathfrak m}$ is the maximal ideal in $C^\infty(M)$ consisting of functions vanishing at $p$. See my answer here. – Moishe Kohan Nov 23 '24 at 04:33
  • The definition from the book on Riemann surfaces is the worst definition of the cotangent bundle that I have ever seen, and I would strongly urge you to forget about it and pick up another book instead. This definition does not generalise in any meaningful way, it only applies to $1$-dimensional manifolds (smooth or complex), and does not provide any meaningful intuition for what the cotangent bundle really is. So no wonder you don't understand the motivation. Shocking that such definitions can be found in publicised textbooks. – Quaere Verum Nov 23 '24 at 16:54
  • Thank you very much @MoisheKohan. I found exactly the same idea from the following blog post, An algebraic definition of the cotangent space – Dr.Antidode Nov 23 '24 at 17:50
  • Yes, it is quite standard, just not everybody knows this. – Moishe Kohan Nov 23 '24 at 17:54
  • Agree with you @QuaereVerum Can you suggest any good resource? Actually, I know differential geometry a little bit. I was preparing a talk for my college seminar. And want to illustrate everything from scratch as well as from different perspectives. For me, I realize understanding tangent and cotangent space is the most challenging part of the DG. that's why want to understand the dualization perspective for each context. – Dr.Antidode Nov 23 '24 at 17:54

1 Answers1

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Here is a terse take on this:

The definition of $T^{\operatorname{glue}}_p$ is based on the fact that local coordinates $(x^1, \dots, x^n)$ induce a basis $(\partial_1, \dots, \partial_n)$ of $T^{\operatorname{glue}}_p$. This in turn induces a dual coordinate basis of the cotangent space, defined as the dual space.

Using the path definition, I would start with a smooth function $f$ and observe that it defines a linear function $df$ on $T^{\operatorname{path}}$ given by $$ df([\gamma]) = (f\circ \gamma)'(0) $$ and define the cotangent space to be the set of equivalence classes. This is essentially the same as the definition $\mathfrak{m}/\mathfrak{m}^2$. This approach also works with $T^{\operatorname{der}}_pM$.

I would suggest that if you are introducing these concepts from scratch to use the definition of a tangent vector as a velocity vector of a curve, observe that the directional derivative of a function is a linear function of the velocity vector, use this to define the differential of a function at a point, and finally define the cotangent space is the space of all possible differentials of functions. I suggest avoding the other two approaches. The derivation approach is an elegant but somewhat opaque way to define the tangent space rigorously. The local trivialization via coordinates approach is messy and confusing.

Deane
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  • Thank you very much @Deane, For $T^{\text{der}*}p$, I guess you mean start with a smooth function $f$ and then we have $X\mapsto Xf$ where $(Xf)(p)=X(p)f=X{\gamma,p}f=(f\circ\gamma)'(0)$, right? – Dr.Antidode Nov 23 '24 at 19:40
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    What I meant about derivations is that given a smooth function $f$, there is a linear function $df: T_p^{\operatorname{der}} \rightarrow \mathbb{R}$, where for each derivation $D$, $$df(D) = D(f). $$ Again, you define the cotangent space to be the set all of such linear functions. – Deane Nov 23 '24 at 19:52
  • I guess I won't get a better answer than yours. Let any $\omega\in T_p^{\text{path}*}$ then I want to understand the local expression for $\omega$ (let for $(z_\alpha,U_\alpha)$). If I understand correctly then $dz_\alpha([\gamma])=(z_\alpha\circ\gamma)'(0)$. For $\omega=\omega^\alpha dz_\alpha$ does it act like: $$\omega([\gamma])\stackrel{?}{=}\omega^\alpha(p)(z_\alpha\circ\gamma)'(0)$$Thanks again for your response @Deane And if you don't mind please enlighten me your pespective to this, like how to understand these in local sense intuitively. I really want to know your's. – Dr.Antidode Nov 28 '24 at 20:30