We define the tangent space at a point, say $x_0$, on the manifold $M$ as the set of all derivations, i.e maps which maps smooth maps from a neighbourhood of $x_0$ to real numbers to real numbers. Then we go and show that the maps $\frac{\partial }{\partial x_i}|_{x_0}$ form a basis for for this (linear) tangent space.

However, to determine whether given two tangent vector $v,w$ at $x_0$ are lin. independent or not, we would need to their decomposition wrt to this basis, and it is in general not easy to find. Therefore, instead, if we can show that $v(f), w(f)$ are lin. independent for all $f$ that is a smooth map from a neighbourhood of $x_0$ to real numbers, we can argue that $v,w$ are lin. independent, but $v(f), w(f)$ are just real numbers, and we cannot check the linear dependent of $v(f), w(f)$ for all $f$, so...


Given two tangent vectors of at a point on the manifold, how do we actually determine whether they are linearly independent or not.


We can consider the base $\{ \frac{\partial }{\partial x^1}, \dots , \frac{\partial}{\partial x^n}\}$ on $T_pM$ induced by the local chart around p, $\phi=(x_1,\dots ,x_n)$. In this case you know that two tangent vector $v,w\in T_pM$ can be written in this way:

$v=v(x_1) \frac{\partial}{\partial x^1}+\dots + v(x_n)\frac{\partial}{\partial x^n}$

$w=w(x_1) \frac{\partial}{\partial x^1}+\dots + w(x_n)\frac{\partial}{\partial x^n}$

$v,w$ will be linearly dependent if and only if there exists $\lambda\neq 0$ such that

$w=\lambda v$

more precisely

$w(x_1) \frac{\partial}{\partial x^1}+\dots + w(x_n)\frac{\partial}{\partial x^n}=\lambda( v(x_1) \frac{\partial}{\partial x^1}+\dots + v(x_n)\frac{\partial}{\partial x^n})$

if and only if

$(w(x_1)-\lambda v(x_1))\frac{\partial }{\partial x_1}+\dots + (w(x_n)-\lambda v(x_n))\frac{\partial}{\partial x_n}=0$


$\{ \frac{\partial }{\partial x^1}, \dots , \frac{\partial}{\partial x^n}\}$

is a base for $T_pM $ so it must be that

$w(x_j)-\lambda v(x_j)=0$ for each $j\in \{1,\dots ,n\}$

So $v,w$ will be linearly dependent if and only if there exists $\lambda\neq 0$ such that

$w(x_j)=\lambda v(x_j)$ for each $j\in \{1,\dots , n\}$


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