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Calculate the Tangent Space of a line $y=mx; m\in \mathbb{R}$ in a point $p=(x,y)$

I know that a line that passes through the origin is a manifold $M$ and the chart is $(M,\varphi); \varphi(x)=(x,mx)$. I know that $T_pM$ is generated by one element, but i don't know how to find the element $d/dx^i$

I appreciate your help.

sango
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2 Answers2

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A line $L:y = mx$ admits a global chart $\mathbb{R} \ni t\mapsto (t,mt) \in L$. The tangent space $T_{(t,mt)}L$ is spanned by $$\frac{\partial}{\partial t}\bigg|_{(t,mt)} = \frac{\partial}{\partial x}\bigg|_{(t,mt)} +m\frac{\partial}{\partial y}\bigg|_{(t,mt)}.$$This is a particular instance of the general fact that the tangent space to a vector space (seen as a manifold) at any point is isomorphic to the vector space itself: if $V$ is a vector space with basis $(v_1,\ldots,v_n)$ and $p\in V$, one takes the global chart $\mathbb{R}^n\ni (x^1,\ldots,x^n)\mapsto \sum_{i=1}^nx^iv_i\in V$, so that the isomorphism is $$T_pV \ni \frac{\partial}{\partial x^i}\bigg|_p\mapsto v_i\in V.$$Here $V = L$ and $v_1 = (1,m)$.

Ivo Terek
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  • what is suppose to be the base of $T_{(t,mt)}L$? – sango Sep 29 '19 at 20:39
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    Please read the answer again carefully, it was essentially the first thing I wrote: the basis consists of the single vector $(\partial/\partial t)|{(t,mt)}$ given there as a specific combination of the canonical coordinate fields of $\Bbb R^2$ (in particular, $T{(t,mt)}L$ is a subspace of $T_{(t,mt)}\Bbb R^2$). Moreover, all identifications work naturally: the isomorphisms $T_pV\cong V$ applied for $V = L$ and $V = \Bbb R^2$ actually take $L\subseteq V$ into $T_{(t,tm)}L \subseteq T_{(t,mt)}\Bbb R^2$ (if you'd like, a certain diagram even commutes). – Ivo Terek Sep 29 '19 at 20:42
  • But isn't the map $\mathbb{R} \ni t\mapsto (t,mt) \in L$ a parameterization, not a chart? To do this carefully, you need to find a slice chart for the line and go from there. But this is easy: the line $l$ is an embedded submanifold with slice chart $(y-mx,y)$ – Matematleta Sep 29 '19 at 20:58
  • Yes, formally what I wrote is a parametrization, and the inverse is the chart. I just followed notation in the original post to avoid confusing OP further. While finding the tangent space as the kernel of some map is definitely more elegant, there is no need to do so. One can use that the coordinate vector fields of a chart ("above to below") form a basis for the tangent spaces, but by definition those can be computed as partial derivatives of the parametrization ("below to above"). – Ivo Terek Sep 29 '19 at 21:02
  • Your answer is fine but there are things to check, right? The OP does not seem to know more than the most basic facts. So, for instance "by definition those can be computed as partial derivatives of the parametrization". Surely not by definition. It follows from the fact that any tangent vector is the derivation (not derivative) of a path $\gamma$ in $l$ s.t. $\gamma(0)=p\in l$ Then, you show that if $\gamma'(0)=a\frac{\partial }{\partial x}+b\frac{\partial }{\partial y}$ then $a=\frac{dx\circ \gamma}{dt}|_{t=0}$ and similarly for $b.$ – Matematleta Sep 29 '19 at 21:16
  • Yes, I'm assuming the very minimum background. If OP wants to go back to the very first definitions of tangent vectors, there is more work to be done. But the point of my answer is that those computations such as finding bases can be done in a "calculus" mindset. – Ivo Terek Sep 29 '19 at 21:19
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We expect the tangent space to be the set of vectors that point in the direction of the line itself. To prove this, note that if $f:\mathbb R^2\to \mathbb R:(x,y)\mapsto y-mx$, then at any $p$ on the line $l,\ T_pl=\text{ker}f_*.$ Now, $f_*\left(a\frac{\partial }{\partial x}+b\frac{\partial }{\partial y}\right)=-am+b$ and this is zero if and only if $b=am$ so $T_pl=\{a\frac{\partial }{\partial x}+am\frac{\partial }{\partial y}:a\in \mathbb R\},$ as expected.

Matematleta
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