# Vectors as a differential op and covectors as differenential

Hey,

I'm new to this , and I understand the derivation of the transition laws for overlapping regions of a manifold for covectors and vectors starting from thinking of them as a differential and a differential operator respectively, but I don't really have a clue where this comes from...

Any assistance or a point toward some good source, greatly appreciated !

Fredrik
Staff Emeritus
Gold Member
There's a more intuitive definition of the tangent space at p that isn't used much, because it's harder to work with. By this definition, a tangent vector at p is an equivalence class of curves through p. The equivalence relation is defined using a coordinate system ##x:U\to\mathbb R^n## such that ##p\in U##. We say that two curves ##C:\mathbb R\to M## and ##B:\mathbb R\to M## such that ##C(0)=B(0)=p## are equivalent if
$$(x\circ C)'(0)=(x\circ B)'(0).$$ It turns out that this definition is independent of the chosen coordinate system. We can then define addition and scalar multiplication on the set of equivalence classes to turn it into a vector space. And then we can prove that the tangent space defined this way is isomorphic to the already familiar tangent space at p. The isomorphism is the map ##\phi## defined by
$$\phi([C])(f)=(f\circ C)'(0)$$ for all equivalence classes [C] and all smooth functions f.

The fact that ##\big((\mathrm dx^1)_p,\dots,(\mathrm dx^n)_p\big)## is the dual of ##\big(\frac{\partial}{\partial x^1}\big|_p,\dots,\frac{\partial}{\partial x^n}\big|_p\big)## is a straighforward consequence of the definition of the ##\mathrm d## operation. The definition is ##(\mathrm df)_p(v)=v(f)## for all smooth ##f:M\to\mathbb R## and all ##v\in T_pM##, so we have
$$(\mathrm dx^i)_p\left(\frac{\partial}{\partial x^j}\bigg|_p\right) = \frac{\partial}{\partial x^j}\bigg|_p x^i =\delta^i_j.$$

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lavinia
Gold Member
Although Fredrik has explained it all, I thought this might be helpful as well.

In ordinary calculus in Euclidean space, vectors operate on functions by directional derivatives. If v is a vector, its action on a function,f, is

lim t ->0 (f( x + tv) - f(x))/t

Conversely one can define vectors at a point,x, as linear operators on functions that obey the Leibniz rule, that is : (v.fg) = (v.f)g(x) + (v.g)f(x)

One can show that such a "derivation" is equal to a directional derivative for some vector.

Similarly df(v) is is defined by the same limit. The way I learned it was that the Jacobian matrix of a function is a linear map between vector spaces. For instance, the gradient of a function is a linear map from n space to the real numbers.The notation df is the invariant way to describe the Jacobian.

As Fredrik explained, on a manifold one only has directional derivatives in a coordinate chart so some notion of equivalence under change of charts is needed. One way to define vectors invariantly without reference to coordinate charts is as equivalence classes of smooth curves at a point, another is as derivations.

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