Why the basis of the tangent space of a manifold is some partials?

[wdlang]

it is quite peculiar

i know you do not want to embed the manifold into a R^n Euclidean space

but still it is too peculiar

it is hard to develop some intuition

 

[quasar987]

Are you really asking why the basis of the tangent space of a manifold is some partials like the title of the thread suggests, or why the tangent space of a manifold is defined the way it is as the vector space of derivations at p on the germs of smooth functions?

 

[wdlang]

Are you really asking why the basis of the tangent space of a manifold is some partials like the title of the thread suggests, or why the tangent space of a manifold is defined the way it is as the vector space of derivations at p on the germs of smooth functions?

i mean the latter

 

[Fredrik]

There's a more intuitive definition. If A and B are curves such that A(0)=B(0)=p, and x is a coordinate system, then we can say that A and B are equivalent if x\circ A and x\circ B have the same derivative at 0. (These are curves in \mathbb R^n, so this doesn't involve differential geometry). Then we define the tangent space at p as the set of equivalence classes of curves that pass through p at parameter value 0. It's easy to show that the equivalence relation doesn't depend on which coordinate system is used in its definition.

That set can be given a vector space structure, again using coordinate systems, and then proving that the definition is independent of the coordinate systems used. If we call that vector space T_pM, and the vector space of derivations D_pM, we can show that the function \phi:T_pM\rightarrow D_pM defined by \phi([A])(f)=(f\circ A)'(0) is a vector space isomorphism. Of course, first we have to prove that it's a well-defined function, by showing that the right-hand side doesn't depend on which member A we choose from the equivalence class [A].

See Isham for details. (It also has a nice proof of the (non-trivial) claim that the set of partial derivative operators associated with a coordinate system is a basis for D_pM).

 

[RedX]

it is quite peculiar

i know you do not want to embed the manifold into a R^n Euclidean space

but still it is too peculiar

it is hard to develop some intuition

The basis doesn't have to be the partials, but the partials are the most natural choice. You can convert between the basis of partials and any other basis by the vielbein.

I'm not sure if this is accurate to say or not, but the basis in classical differential geometry is:

e_i=\frac{\partial \vec_{r}}{\partial x^i}

where r is the position vector in a euclidean embedding space. modern differential geometry just gets rid of the r:

e_i=\frac{\partial}{\partial x^i}

 

[Bacle]

I don't know if this is the type of answer that you are looking for, but, following
up on Fredrik's post, you can show that every derivation is a partial derivative,
(and the fact that the space of derivations has the same dimension as the
manifold in which it lives) so that the partial derivatives, as derivations/operators,
span the tangent space. Sorry, my knowledge of physics is too limited to give
you a physical explanation.

 

[wdlang]

There's a more intuitive definition. If A and B are curves such that A(0)=B(0)=p, and x is a coordinate system, then we can say that A and B are equivalent if x\circ A and x\circ B have the same derivative at 0. (These are curves in \mathbb R^n, so this doesn't involve differential geometry). Then we define the tangent space at p as the set of equivalence classes of curves that pass through p at parameter value 0. It's easy to show that the equivalence relation doesn't depend on which coordinate system is used in its definition.

That set can be given a vector space structure, again using coordinate systems, and then proving that the definition is independent of the coordinate systems used. If we call that vector space T_pM, and the vector space of derivations D_pM, we can show that the function \phi:T_pM\rightarrow D_pM defined by \phi([A])(f)=(f\circ A)'(0) is a vector space isomorphism. Of course, first we have to prove that it's a well-defined function, by showing that the right-hand side doesn't depend on which member A we choose from the equivalence class [A].

See Isham for details. (It also has a nice proof of the (non-trivial) claim that the set of partial derivative operators associated with a coordinate system is a basis for D_pM).

i understand this literally

but i mean it is a bit peculiar, it is non-intuitive

maybe i need to get accustomed to this

 

[quasar987]

i mean the latter

The most difficult part is perhaps to identify what it mean to define a notion of tangent space for an abstract manifold.

We have the notion of the tangent space for a k-manifold M embedded in some R^n: if \sigma:U\subset \mathbb{R}^k\rightarrow \mathbb{R}^n is a local parametrization of M with \sigma(0)=p, then T_pM is defined as the k-dimensional subspace \mathrm{Im}(D\sigma_0). A basis for it is \left(\frac{\partial\sigma/}{\partial x^i}(0)\right), and if \tau:V\subset \mathbb{R}^k\rightarrow \mathbb{R}^n is another local parametrization of M with \tau(0)=p, then by the chain rule, the relation between the basis \left(\frac{\partial\sigma}{\partial x^i}(0)\right) and \left(\frac{\partial\tau}{\partial x^i}(0)\right) is

\frac{\partial\tau}{\partial x^i}(0)=\sum_{j=1}^k\frac{\partial(\sigma^{-1}\circ\tau)^j}{\partial x^i}(0)\frac{\partial\sigma}{\partial x^j}(0)

My guess then is that the "tangent space problem" is (or rather, was to the pioneers of the theory of calculus on manifolds) a way to associate a k-dimensional vector space T_pM at every point p of an abstract smooth k-manifold M in such a way that each local parametrization of M induces a basis in such a way that if (v_1,...,v_k) and (w_1,...,w_k) are two basis induced by local parametrization \sigma and \tau, then

w_i=\sum_{j=1}^k\frac{\partial(\sigma^{-1}\circ\tau)^j}{\partial x^i}(0)v_i

(compare with the above relation for the classical tangent space)

Indeed, if this program is carried out succesfully, then we will have found way to associate a vector space to each point of a manifold that coincides for manifold embedded in R^n with the classical notion of the tangent space (up to the canonical isomorphism v_i\leftrightarrow \partial\sigma/\partial x^i).

There are at least 3 substantially different ways to solve the tangent space problem, all of which are pretty much equally abstract and initially unintuitive. One of which is the "derivation at p" solution.

 

[quasar987]

i understand this literally

but i mean it is a bit peculiar, it is non-intuitive

maybe i need to get accustomed to this

It does seem to get better with time :)

 

[Rasalhague]

See Isham for details.

I've only dipped into this so far, so there may be an explanation in there that I haven't got to yet... but in Isham's definition 2.4, pp. 73-74, part 2, he writes

Two curves are tangent at a point p in M if

(a) \enspace \sigma_1(0)=\sigma_2(0)=p

(b) In some local coordinate system around the point, the two curves are 'tangent' in the usual sense as curves in \mathbb{R}^m:

\frac{\mathrm{d} x^i}{\mathrm{d} t}(\sigma_1(t=0))= \frac{\mathrm{d} x^i}{\mathrm{d} t}(\sigma_2(t=0))

for i=1,2,...,m (eq. 2.3.2).

(I changed the notation slightly in eq. 2.3.2, as I don't know how to do that full-length vertical line in LaTex.)

My question: suppose we have a curve in the Euclidean plane, such as a circle, paramaterized in orthonormal coordinates by (\cos(t),\sin(t)), t \in [0,2pi), and another curve, say (\cos(t),2\sin(t)), t \in [0,2pi). Would these curves be called tangent at t=0? I would expect yes, as the derivatives are parallel here, but the derivative of the latter curve with respect to t, at this point, has twice the magnitude of the derivative of the unit circle, and so seems not to meet condition (b). Does this definition assume that t is arc length so that any curve has unit speed?

 

[Fredrik]

I don't think I even thought of that when I read the book. I don't really have time to think about it now, but if you figure it out on your own, let me know.

You can use \bigg| to get a result like this:

\frac{d}{dt}\bigg|_0 f(t)=f'(0)

 

[llarsen]

Here is my shot at explaining why partials are used as the basis for the tangent space. First it is useful to think a little bit about the manifold itself:

There are two related approaches that can be used to describe a manifold. We can represent it in terms of a set of n coordinate functions, where n is the dimension of the manifold (such as x, y, and z in euclidean 3 space). The coordinate functions may be embedded in a larger space (such as functions to describe the 2D surface of a sphere in 3D space).

Another way we tend to represent a manifold is in terms of coordinate lines (i.e. parameterized coordinate curves). The coordinate curves can be derived from the set of coordinate functions, and for each coordinate function, there is a corresponding set of coordinate curves. If you draw a picture of the coordinate functions, you can visualize the coordinate curves. For example, in polar coordinates, the coordinate curves associated with the coordinate r are drawn along lines of constant \theta. The parameterization of the coordinate lines is defined by the function r, or in other words, the rate of change of the coordinate curve parameter exactly agrees with the rate of change of the function r. The coordinate curve can be written as:

[x,y] = [r cos(\theta_0), r sin(\theta_0)]

Since this is a one dimensional curve, \theta_0 is held constant while r varies. The associated family of curves is obtained by adjusting the constant \theta_0. The coordinate curve for \theta is given by:

[x,y] = [r_0 cos(\theta), r_0 sin(\theta)]

where r_0 is held constant.

The coordinate functions and coordinate curves each define a basis for the manifold, and are referred to as duals of one another, partially because if one is given, the other can be derived. A basis for the tangent space can be derived by taking the derivative of the coordinate line with respect to the curve parameter. For the r coordinate curve above, the curve parameter is r, and we get a basis vector:

[\frac{dx}{dr}, \frac{dy}{dr}] = [cos(\theta_0), sin(\theta_0)]

Another way to write this is the partial notation (that comes from the chain rule):

\frac{d}{dr} = \frac{dx}{dr}\frac{\partial}{\partial x} + \frac{dy}{dr}\frac{\partial}{\partial y} = cos(\theta_0) \frac{\partial}{\partial x} + sin(\theta_0) \frac{\partial}{\partial y}

So, why the partial notation? Since the coordinate curve for r is a parameterized curve in space, I can use it to take a directional derivative. That is, I can overlay a function from space onto the curve and compare the rate of change of the function compared to the curve parameter (i.e. I can calculate \frac{df}{dr}). I can do this by applying \frac{d}{dr} as defined above to some function of x and y. For example if f=x then:

\frac{df}{dr} = cos(\theta_0) \frac{\partial(x)}{\partial x} + sin(\theta_0) \frac{\partial (x)}{\partial y} = cos(\theta_0)

The second basis tangent vector comes from the equation for the \theta curve, and comes again by using the chain rule:

\frac{d}{d \theta} = \frac{dx}{d \theta}\frac{\partial }{\partial x} + \frac{dy}{d \theta}\frac{\partial}{\partial y} = - r_0 sin(\theta) \frac{\partial}{\partial x} + r_0 cos(\theta) \frac{\partial}{\partial y}

So in general, the tangent basis can be derived once a coordinate chart is chosen, and the coordinate curves are calculated. The tangent basis can be thought of a set of directional derivatives along the coordinate curves with respect to the curve parameter (which is defined by the associated coordinate function). It is a way of defining derivatives with respect to the selected coordinate functions.

 

[Phrak]

The following is something that is not exact in so far as ignoring the tangle of maps, but may be an intuitive start to the graphically minded. The action of a vector on a dual vector is the scalar df/d\lambda. This is the action of a directional derivative on a scalar valued function.

\frac{df}{d\lambda} = \frac{\partial f}{\partial x^i}\frac{dx^i}{d \lambda}

\frac{df}{d\lambda} = \frac{\partial f}{\partial x^i}\ \frac{\partial x^i}{\partial x^j} \frac{dx^j}{d \lambda}

\frac{df}{d\lambda} = \frac{\partial f}{\partial x^i}\ \partial_j dx^i \ \frac{dx^j}{d \lambda}
​

We will demand that

\partial_j dx^i} = \delta^i_j​

for the reason that follows. We can rearrange, and partition the right side of the equation, so long as we remember that the action of \partial_j is only upon dx^i.

\frac{df}{d\lambda} = \left[ \frac{dx^j}{d \lambda} \partial_j \right] \left[ \frac{\partial f}{\partial x^i} dx^i \right]
​

This,

V^j \hat{e}_j = \left[ \frac{dx^j} {d \lambda} \partial_j \right]​

is a vector with basis \partial_j acting upon the dual vector, or covector, or one-form (or anyway you find it named),

U_i \hat{e}^i = \left[ \frac{\partial f}{\partial x^i} dx^i \right]​

with bais dx^i.

The demand that

\partial_j dx^i} = \delta^i_j​

is exactly the requirement of dual vector basis.

The action of a vector on a one-from is the action of the vector basis upon the one-form basis, where the vector and dual vector coefficents act though direct multiplication.

 

[llarsen]

One thing you will begin to notice is that the chain rule plays a fundamental role in differential geometry. In differential geometry, we don't want to restrict ourselves to the viewpoint of a single coordinate system, and the chain rule is the rule of transforming between coordinate systems. Because of this it is useful to represent differential objects in a notation that is natural for the chain rule.

It does seem a little disconcerting to use partials as basis vectors at first, particularly if you are thinking of a tangent basis at a point on the manifold. This notation is more naturally suited to expressing vector fields that fill the manifold rather the basis at a point.

You may remember that an ordinary differential equation can be expressed in first order form (i.e. a form in which no derivative higher than first order appears), and that this can be represented as a vector field on state space. Commonly ordinary differential equations are represented in the notation:

\frac {dx_1}{dt} = f_1(x), \frac {dx_2}{dt} = f_2(x),..., \frac {dx_n}{dt} = f_n(x)

It is trivial to convert this to vector field form when you are using partial notation, because the vector field form looks just like the chain rule (on purpose):

\frac{d}{dt} = \frac{dx_1}{dt} \frac{\partial}{\partial x_1} + \frac{dx_2}{dt} \frac{\partial}{\partial x_2} + ... + \frac{dx_n}{dt} \frac{\partial}{\partial x_n} \\ <br /> = f_1(x) \frac{\partial}{\partial x_1} + f_2(x) \frac{\partial}{\partial x_2} + ... + f_n(x) \frac{\partial}{\partial x_n}

When the differential equation is written in this form, the rule for transforming the differential equation to a new coordinate system becomes obvious. Suppose we would like to transform to a new coordinate system y, then the differential equation can be written as:

\frac{d}{dt} = \frac{dx_1}{dt} \frac{\partial y_i}{\partial x_1} \frac{\partial}{\partial y_i} + \frac{dx_2}{dt} \frac{\partial y_i}{\partial x_2} \frac{\partial}{\partial y_i} + ... + \frac{dx_n}{dt} \frac{\partial y_i}{\partial x_n} \frac{\partial}{\partial y_i}

Or this can be represented in the more compact notation:

\frac{d}{dt} = \frac{dx_j}{dt} \frac{\partial y_i}{\partial x_j} \frac{\partial}{\partial y_i}

Note that the differential equation can be integrated, and that the solution curves are paratmeterized lines that fill the space. The derivative of the solution curves is the differential equation, or the vector field. The vector field can be used as a derivative operator (the Lie derivative), which has a natural interpretation as comparing a function q with the solution parameter t (i.e. \frac{dq}{dt}) along the solution curve of the differential equation. Differential geometry is filled with chain rule operations when dealing with differential objects (such as ODEs or tensor fields), so the partial notation for a basis of the tangent space is natural and convenient when working with these kind of objects.