21 votes
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Why aren't tangent spaces simply defined as vector spaces with same dimension as the manifold?

We don't just want to have a vector space to call the "tangent space". We want to do geometric things with the tangent space, and we can't do those things if it's just an arbitrary vector space of ...
Eric Wofsey's user avatar
18 votes
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Lie algebra: intuition of "Lie Algebra is tangent space of corresponding Lie Group"?

Intuitively, you can think of the tangent space of a surface at a point as the space of all "directions" you can move from that point (with all velocities) while staying on the surface. That ...
Toffomat's user avatar
  • 2,225
16 votes

Why are the partial derivatives a basis of the tangent space?

To distinguish between points and tangent vectors, let $p=(p_1,...,p_n)\in \mathbb{R}^n$ a point of $\mathbb{R}^n$ and $v=(v_1,...,v_n)\in T_p(\mathbb{R}^n)$ a point of the tangent space $\mathbb{R}^n$...
bing-nagata-smirnov's user avatar
13 votes

Why aren't tangent spaces simply defined as vector spaces with same dimension as the manifold?

If they are “just vector spaces” at each point, then every set would be a manifold. The whole point is to link the topological structure of the manifold (using coordinate patches that fit together ...
rschwieb's user avatar
  • 153k
12 votes
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Why is the tangent bundle defined using a disjoint union?

You are right, the "usual" constructions of $T_p M$ as a set of derivations or a set of equivalence classes of curves produces disjoint tangent spaces at distinct points. Thus you could define $T'M = \...
Paul Frost's user avatar
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12 votes

Why is the tangent bundle defined using a disjoint union?

In my book Introduction to Smooth Manifolds, I define $T_pM$ as the set of all linear maps $v\colon C^\infty(M)\to\mathbb R$ that satisfy $$ v(fg) = f(p)vg + g(p)vf $$ for all $f,g\in C^\infty(M)$. ...
Jack Lee's user avatar
  • 46.3k
11 votes
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How exactly does it make sense to differentiate a function whose input is a point on a manifold?

Very clearly $(\partial f\circ x^{-1})x(p)$ is not what was written on the board; I assume you just made a typo. What was written is $[\partial_i(f\circ x^{-1})](x(p))$. This is simply the $i^{th}$ ...
peek-a-boo's user avatar
  • 54.2k
10 votes
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Motivation for defining a tangent vector as an equivalence class of curves

A few preliminary remarks about the definition: there is actually no need for $\alpha$ to be defined on $(-1,1)$; all you need is for $\alpha$ to be defined on a small interval $(-\varepsilon, \...
peek-a-boo's user avatar
  • 54.2k
10 votes
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Understanding vector field(s) on $\mathbb{S}^3$.

$S^3$ is often defined as a subset (submanifold) of $\Bbb{R}^4$, so at a point $p\in S^3\subset \Bbb{R}^4$, the tangent space $T_pS^3$ is a subset of $T_p\Bbb{R}^4$ (or atleast there is a very natural ...
peek-a-boo's user avatar
  • 54.2k
9 votes

$T_pS \subseteq T_pM$: Are tangent spaces of submanifolds subsets (and not embedded) of tangent spaces of the original manifold?

I believe both equalities are really equalities are not isomorphisms This is not quite true; it might help to write out the definitions. Given a manifold $M,$ we can form the vector space $C^{\infty}(...
ktoi's user avatar
  • 7,247
9 votes
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Why is $\mathfrak{m}/\mathfrak{m}^{2}$ the cotangent space to a variety?

Your intuition is not correct: a tangent vector at a point is something you can take a directional derivative along. This leads naturally to the identification of the tangent space with the ...
KReiser's user avatar
  • 64.1k
7 votes
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Basis of the space of vector fields on smooth manifolds

The set of vector fields of a manifold $M$ is infinite dimensional. For example, over $\mathbb{R}$, it is isomorphic to $\mathcal{C}^{\infty}(\mathbb{R})$ because the tangent bundle is trivial, and ...
Didier's user avatar
  • 18.9k
7 votes
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Tangent space of a point

A single point is $0$-dimensional, so the tangent space should also be a zero dimensional vector space. Similarly, you'll only have $\Omega^0(M)$ being nonzero ($\Omega^i(M)=0$ for $i>0$). These ...
Michael Burr's user avatar
  • 32.8k
6 votes

Coordinate basis and coordinate systems

This is not a proper answer to this post, but it might help the OP find out the answers they seek anyway. So, if we're doing differential geometry and we're considering the tangent space $T_P\mathcal{...
TeicDaun's user avatar
  • 192
6 votes
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Differential of Multiplication Map of Lie Group

First note that the inverse of the isomorphism $\alpha$ is the map $\tau :T_pM \oplus T_qN \longrightarrow T_{(p,q)}(M\times N)$ defined as $$ \tau(v,w)=d\iota_1(v)+d\iota_2(w), $$ where $\iota_1 : M ...
Kelvin Lois's user avatar
  • 6,990
6 votes
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Krull's Principal Ideal Theorem for tangent spaces

$f$ gives an element of the cotangent space $m/m^2$. Since the tangent space is the dual of the cotangent space, we can evaluate elements of the tangent space on $f$. "The tangent space of $A/f$ is ...
KReiser's user avatar
  • 64.1k
6 votes
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Basis for the Tangent space and derivations at a point

$\newcommand{\R}{\mathbb{R}}$ So because we are in $\mathbb{R}^n$ we can all imagine (or at least were told) how we should visualize what a tangent space is at a certain point $p$ in $\mathbb{R}^n$, i....
hal4math's user avatar
  • 1,120
6 votes
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Canonical/tautological 1-form

A $1$-form $\lambda$ on $T^*Q$ is a section of the cotangent bundle of $T^*Q$, so a map $\lambda:T^*Q\to T^*(T^*Q)$. So $\lambda$ has first to take a point $x\in T^*Q$ (with $\pi(x)=q$, which means ...
Balloon's user avatar
  • 8,344
6 votes
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Tangent space of a union of closed affine varieties

Let $X_1 = V(y)$ and $X_2 = V(y-x^2)$. Then the tangent spaces at zero of both these varieties is the $x$-axis. But their union is the variety $W := V(y(y-x^2))$. Then as the defining polynomial ...
James's user avatar
  • 1,575
6 votes
Accepted

Tangent space definition

Suppose $U$ and $V$ are two neighborhoods of $p$, $f$ is defined on $U$ and $g$ is defined on $V$. How do you define $f+g$? (You do need at least an abelian group structure if you want to define ...
Captain Lama's user avatar
  • 25.3k
6 votes
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The Universal Property of Tangent Spaces

$\newcommand{\dif}{\mathrm{d}}$ The following is inspired by the pages 362-3 of Tammo tom Dieck's book "Algebraic Topology". The definition of tangent spaces and the differential exploits ...
Filippo's user avatar
  • 3,402
6 votes

How exactly does it make sense to differentiate a function whose input is a point on a manifold?

Let $M$ be an $n$-dimensional manifold, $p \in M$ a point, $x: U \subset M \to x(U) \subset\mathbb{R}^n$ a chart around $p$ and $f: V \subset M \to \mathbb{R}$ a function such that $p \in V$ (with $V$ ...
Matheus Andrade's user avatar
5 votes
Accepted

What is the motivation of creating of $T^*_p(\mathbb R^n)?$ How can we visualize covectors?

I only given an answer to the first question. My answer is an addition to the answer given by janmarqz. Basically I just provide some plots for his second point. Visualization I like the ...
Steffen Plunder's user avatar
5 votes
Accepted

Tangent space on the sphere

You are almost there for $\subset$ since $f(x)\in S^m$, $\langle f(x),f(x)\rangle =1$ so it is constant, and for the other direction, use the fact that the dimensions of the image $df_x$ and $TS^m_p$ ...
Tsemo Aristide's user avatar
5 votes
Accepted

Isn't this a trivial corollary?

Take$$M=\{(x,y)\in\mathbb R^2\mid x^2+y^2=1\}$$and let $f(x,y)=x^2+y^2$. Then $f|_M$ is constant, and therefore it has an extreme point at every point of $M$. However, $\nabla f$ is never equal to $0$,...
José Carlos Santos's user avatar
5 votes

Coadjoint action of Lie group on dual of its Lie algebra

This definition is compressing different steps together and IMO it's cleaner to separate them, especially if e.g. taking duals of representations is unfamiliar. The dual linear map In general, if ...
Qiaochu Yuan's user avatar
5 votes
Accepted

1-forms vs k-forms vs Vector Fields

A section of $\bigwedge^kTM$ is called a polyvector field (or multivector field) of degree $k$. Just as you can take the wedge product of differential forms, you can take the wedge product of ...
Michael Albanese's user avatar
5 votes

Different definitions of the tangent space at a point of a smooth manifold: Biduals?

We're talking about finite-dimensional real vector spaces, so a posteriori $T_p M$ has to be the dual of $T^*_p M$ and $T^*_p M$ has to be the dual of $T_p M$. I'm just mentioning it to emphasize that ...
Michał Miśkiewicz's user avatar
5 votes
Accepted

The derivative of a smooth map between smooth submanifolds of Euclidean spaces: Is it a best linear approximation?

Yes, it's correct, except for the technical point that the derivative map acts on (and maps to) the tangent space, not the affine tangent space. The tangent vector $u'(0)$ is based at the origin, not ...
Ted Shifrin's user avatar
5 votes
Accepted

When defining tangent spaces, given $\gamma:(-1,1)\to M$ and $\varphi:U\to\mathbb R^n,U\subseteq M$, why is $\varphi\circ\gamma:(-1,1)\to\mathbb R^n$?

This is marginally imprecise as written. For the composition to make sense, we have to have $\gamma_1((-1,1)),\gamma_2((-1,1))\subseteq U$. However, note that $\gamma_1(0)=\gamma_2(0)=x\in U$, so, by ...
Thorgott's user avatar
  • 10.9k

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