Visualizing the cotangent space to a sphere

[giulio_hep]

TL;DR Summary

Comparison of tangent and cotangent space to a point of a sphere

Is it correct to say that:
the cotangent is given by the gradients (*) to all the curves passing through a point and it actually spans the same tangent space to a point of a sphere? If you visualize them as geometric planes (**), the cotangent and the tangent spaces are more than isomorphic, they're the same plane, at least in this example, forget the co-vector convention?

(*) see for example
> A cotangent vector can be thought of as a gradient.
from http://math.ucr.edu/home/baez/gr/cotangent.vector.html

(**) for the tangent space,
please see the pictorial representation built in this video, especially at the 27 minute captured in the link.
For the cotangent space, I'm trying to find a similar geometric interpretation.

 

[giulio_hep]

Ok, so I've somehow, geometrically, identified the tangent space with the tangent plane.
Now I think that the answer is that, yes, I can represent the dual vector space, the cotangent space, which is also a one-form and so can be visualized with helpful pictures, as suggested in a similar thread of another forum, with the same pictorial tangent plane, by selecting a chart coordinate function, such that the cotangent vector basis is perpendicular to the tangent vector basis but belonging to the same tangent plane, as well as I can imagine a gradient of a curve as perpendicular to the tangent (of the curve as level set) at a specific point.

 

[zinq]

When a vector space V over the real numbers R is endowed with the additional structure of an inner product (a positive definite bilinear mapping B : V x V → R), then there is a natural isomorphism between the vector space and its dual space V* (the real vector space of all linear maps L : V → R). This is given by a function f : V → V* defined as follows: For any vector v in V, f(v) is the linear mapping L : V → R defined by L(w) = B(v,w). (Or in more familiar terms: the vector v gets taken to "dot product with v".

(I used to think this duality was useless and boring, but it isn't.)

OK, so assuming you're speaking of a *Riemannian* manifold M that is a sphere, such as the unit sphere S² in 3-space (you didn't say which sphere you are referring to), then for each x in M, the tangent plane T_xM to M at x has a well-defined inner product. If specifically M = S² in 3-space then vectors in the tangent space T_xM = T_xS² get their inner product from the restriction of the Eucidean on on R³.

But don't just say "the cotangent" as in the original question. There are cotangent vectors, which at a given point of the manifold make up the cotangent plane at that point, and the disjoint union of all the cotangent planes of a manifold form the cotangent space of the manifold. (Often denoted as T*M.) So it's important to specify what you mean.

And you seem to be confused about what a 1-form is. A 1-form ω on a manifold is a choice of a cotangent vector at each point of the manifold. In other words for each x in M, ω(x) is a cotangent vector, belonging to the cotangent space at x. (And, this must be done so that the choice varies smoothly from one x to another.) If you know what a "section of a bundle is": a 1-form is a smooth section of the cotangent bundle.

You can think of a 1-form as a creature that eats vector fields and spits out real-valued functions. For, if ω is a 1-form on the manifold M, and U is a vector field on M (for each x in M, a smooth choice U(x) of tangent vector in the tangent space at x), then by plugging in U(x) to ω(x), we get ω(x)(U(x)), which you can check is just a real number for each x in M. All these real numbers then define a smooth real-valued function on M.

 

[giulio_hep]

I (and wikipedia, etc..) call $$T^*M$$ the cotangent bundle, whilst I identify its section at x as $$T^*_xM$$ i.e. the cotangent space/plane, formed by all the cotangent vectors at point x.

 

[Orodruin]

I (and wikipedia, etc..) call $$T^*M$$ the cotangent bundle, whilst I identify its section at x as $$T^*_xM$$ i.e. the cotangent space/plane, formed by all the cotangent vectors at point x.

This is not correct nomenclature. A section of a fibre bundle $E$ over a manifold $M$ is a map $\sigma : M \to E$ such that $\pi \circ \sigma$ (where $\pi$ is the projection from $E$ to $M$) is the identity map on $M$. In other words, it is a map from a point $x$ to the fibre at $x$.

A section of the cotangent bundle is therefore the assignment of a single dual vector in $T^*_x M$ to each point $x$ in $M$. It is not the same thing as the set of all dual vectors at a single point.

 

[giulio_hep]

This is not correct nomenclature.

Before going in deeper details, please let me ask the following:
@Orodruin Are you only referring to the term "section" in my previous reply?
In other words, as a preliminary clarification (before I dive into the fibers and the base space...), do you confirm/agree with the nomenclature of cotangent bundle as $$T^*M$$ and cotangent space as $$T^*_xM$$ and with the fact that they are different?

 

[Orodruin]

Before going in deeper details, please let me ask the following:
@Orodruin Are you only referring to the term "section" in my previous reply?
In other words, as a preliminary clarification (before I dive into the fibers and the base space...), do you confirm/agree with the nomenclature of cotangent bundle as $$T^*M$$ and cotangent space as $$T^*_xM$$ and with the fact that they are different?

Yes, the cotangent bundle is the union of all cotangent spaces, $T^*M = \bigcup_{x\in M} T_x^*M$.

To be a bit more specific, $T_x^*M$ is the cotangent space at x.

 

[giulio_hep]

Ok thank you. I agree with your definition of "section", sorry for the confusion.
Having said that and back to my original question, can we also confirm that the tangent plane at x and the cotangent plane at x are the same plane, geometrically?

 

[giulio_hep]

@zinq thanks, by the way I was listening to a lection of Frederic Schuller on YouTube (minute 15:25) about the cotangent vector as gradient and he was saying something like "... let's assume I don't provide this inner product .." and " ... I don't need inner product because I can simply say ... ", "... Let x be tangent to a level set ... ". I'm not completely sure I fully understand: is it clear to you what he says, is the inner product maybe not needed in this case?

 

[Infrared]

Let $S=\{f=c\}$ be a level set of $f$ and let $X$ be a tangent vector at some point $x\in C$. Then, a priori, the claim that $\langle\nabla f(x),X\rangle=0$ depends on a choice of a metric. But in fact, unwinding the definitions, it is just saying that $df(X)=0$, which is obviously true (and independent of a choice of metric). This is all he is saying.

 

[giulio_hep]

Ok, in the meantime I've also found another answered question about a geometric representation of covectors (that sounds like an alternative way of asking my question and like answering it with a different picture for the covector space than a simple geometric plane) so I guess that, maybe, the whole topic is now clear enough to me.

 

[giulio_hep]

But in fact, unwinding the definitions, it is just saying that $df(X)=0$, which is obviously true (and independent of a choice of metric)

Well, to be honest I've tried to unwind the definitions he gave before, but I fail to see how it is mathematically proved and I could not find a formal explanation. I'm sorry to ask since you say it's obviously true, but I still find some kind of contradictions because, at start, he also seems to admit that there is no canonical isomorphism between tangent and cotangent spaces without some extra structure (so a metric imho)...

This is what I see: he is probably using the exterior product to define $df$ (but is it part of a vector space without extra structure?) instead of the inner product and it is zero for a vector X acting on itself.
So - as far as the exterior product doesn't require extra structure added to a bare vector space (but I'm not sure, is this point correct?) - a musical isomorphism and/or metric are not needed in the above proof.

Or maybe the correct conclusion is that the above proof requires a metric but it is independent of the choice of a metric and that the gradient definition requires an exterior product but not an inner product?

 

[zinq]

"... the claim that ⟨∇f(x), X⟩ = 0 depends on a choice of a metric."

I would go even further: The gradient vector field ∇f (which equals ∇f(x) at the point x) is *defined* only in case there is a choice of Riemannian metric on the manifold in question.

(And to be ultra-specific: The expression ⟨∇f(x), X⟩ means ⟨∇f(x), X(x)⟩, since the vector field X is evaluated at the point x before being plugged into the inner product.)

In that case, there is a natural isomorphism between the tangent space T(M) and the cotangent space T^*(M) as in #3, namely, the vector v in T(M) corresponds to the dual vector ⟨v, ⟩. (I hope it's clear how this dual vector is defined!)

The "gradient vector field" denoted by ∇f(x) is then defined as the *inverse* of this isomorphism, applied to the 1-form df at each point. So at each point x, there is a dual vector df(x). Then ∇f(x) is the vector at x defined by the condition that "dotting any vector w with ∇f(x)" gives the same answer as applying the dual vector df(x) to w. (Here of course w lies in Tx(M).) We could denote the result by df(x)(w) but usually in notation the x is suppressed so it reads df(w).

So if we want to say ⟨∇f(x), X(x)⟩ = 0 (when ∇f(x) is defined in the presence of a metric), that is the same as saying that at each x, when the dual vector df(x) is applied to the vector X(x), the result is 0.

But applying df(x) to any vector v (at x) means the same thing as the directional derivative of the function f in the direction of the vector v (all taking place at x).

When is that equal to 0 ? Exactly when v is tangent to a level surface of the function f.

And this condition does not depend on the Riemannian metric. Which I believe is what Frederic Schuller was getting at as quoted in #9.

(Note: If ⟨∇f(x), X(x)⟩ equals a nonzero constant c, this condition ⟨∇f(x), X(x)⟩ = c does depend on the Riemannian metric.)

 

[Infrared]

When is that equal to 0 ? Exactly when v is tangent to a level surface of the function f.

This isn't true. Let $M=\mathbb{R}$ and $f(x)=x^2$. Then the gradient of $f$ vanishes at the origin, but $f$ doesn't have level sets of any positive dimension.

(Note: If ⟨∇f(x), X(x)⟩ equals a nonzero constant c, this condition ⟨∇f(x), X(x)⟩ = c does depend on the Riemannian metric.)

Why? The statement $df(X)=c$ doesn't involve any metrics.

 

[zinq]

To respond:

1) I am not qualifying the term "level surface of the function f" by whether it is a bona fide surface. As I use the term, it just means the inverse image f^-1(c) by f of some constant c:

f^-1(c) = {x ∈ M | f(x) = c}.

2) I wrote "this condition does depend on the Riemannian metric." Of course, as you say, the condition df(X) = c does not depend on any choice of metric. And that is equivalent to saying that ⟨∇f(x), X(x)⟩ = c. So in that sense you are of course right.

But (and I see that I expressed this unclearly) the meaning of the vector ∇f(x) at x does depend on the choice of metric; with a different metric it would be in general be a different vector, also denoted ∇f(x), whose inner product with some vector X at x that equals the constant c.

If ⟨∇f(x), X(x)⟩ = c is true with one choice of metric, then it is true in another choice of metric. But by changing the metric we have in general changed both the vector symbolized by ∇f(x), and of course the inner product itself.
So the statement remains true, but the identities of two of its major characters — the gradient vector and the inner product — are different.

 

[Infrared]

1) I am not qualifying the term "level surface of the function f" by whether it is a bona fide surface. As I use the term, it just means the inverse image f^-1(c) by f of some constant c:

f^-1(c) = {x ∈ M | f(x) = c}.

I understand that. It is not true. In my example, $\langle \nabla f(0),V\rangle=0$ for every $V\in T_0\mathbb{R}$ even though $V$ is not tangent to any level set (since any level set is $0$-dimensional and doesn't have nonzero tangent vectors).

If you want a less trivial example, let $f:\mathbb{R}^2\to\mathbb{R}$ be the function $f(x,y)=x^2.$ The level set $\{f=0\}$ is the line $x=0$. The vector $V=\langle 1,1\rangle$ (say based at the origin) is not tangent to this level set, but nevertheless $\langle\nabla f(0,0),V\rangle=0$ because the gradient vanishes.

It is true that if $V$ is tangent to a level set of $f$, then $\langle\nabla f,V\rangle=0$, but the converse just fails.

 

[zinq]

I think we may have differing definitions of tangency.

 

[Infrared]

I just mean that it is in the tangent space of the level set. Can you clarify what you mean?

To respond:

But (and I see that I expressed this unclearly) the meaning of the vector ∇f(x) at x does depend on the choice of metric; with a different metric it would be in general be a different vector, also denoted ∇f(x), whose inner product with some vector X at x that equals the constant c.

If ⟨∇f(x), X(x)⟩ = c is true with one choice of metric, then it is true in another choice of metric. But by changing the metric we have in general changed both the vector symbolized by ∇f(x), and of course the inner product itself.
So the statement remains true, but the identities of two of its major characters — the gradient vector and the inner product — are different.

Okay, but this is true whether or not $c=0$. What is the distinction between these cases that you mean to make? What is metric-invariant if $c=0$ that is not when $c\neq 0$?

 

[giulio_hep]

At this point I'm confused.
Please can someone explain the following definition to me? Is there any reference where it is reported?
The map ${d_P} : C^\infty(M) \mapsto T^*_PM$ would be defined
$$({df})(x) := Xf$$ with vector $x \in T_pM$ in the tangent space.
What is exactly $Xf$ ("the action of the vector field on the provided smooth function") here?

 

[Infrared]

Yes, any introduction to manifolds text should explain it. One way to define the tangent space $T_pM$ is as the vector space of derivations at $p$ (a derivation at $p$ is an $\mathbb{R}$-linear map $X_p:C^{\infty}(M,\mathbb{R})\to\mathbb{R}$ satisfying the Leibniz rule $X_p(fg)=X_p(f)g(p)+f(p)X_p(g)$).

Then $X_p(f)$ is just a number, and we can define $df(X_p)$ to be this number.

 

[giulio_hep]

To define the directional derivative as the action of a vector on a one-form, they choose a curve passing through a point and having that tangent vector and then they compute the ordinary derivative of the curve composed with the function. By the way, they have to prove that the result is not dependent by the specific curve.
Still the definition of directional derivative seems to be based on the concept of a curve initialized by a tangent vector at a point. So it remains very confusing how the notion of tangent space can be also based, in turn, on a derivation, already defined using the concept of tangent vector. It all sounds auto referential and not really constructive...

 

[Infrared]

derivation, already defined using the concept of tangent vector

I gave a definition of a (point) derivation above, which didn't reference any pre-existing idea of tangent vectors.

There are many equivalent definitions of the tangent space. I like the derivation one. Sometimes it's instead defined by equivalence classes of curves, as you're referring to. If $X$ is the equivalence class of a curve $\gamma(t)$, then you can define $df(X)=X(f)=\frac{d}{dt}\big\vert_{t=0}f(\gamma(t)).$

 

[giulio_hep]

Afaics, at #20 you defined the Leibniz rule for a map $X_p$ without showing a constructive method to build such $X_p$: I didn't consider it a complete definition of $X_p$ but only a hint, so I searched for a continuation... but sure, in pure #20, you didn't reference a pre-existing idea of tangent vectors: that happened when I tried to find an "operative" definition.

 

[Infrared]

I assume you mean $X$ instead of $p$ (as $p$ is just a point in a manifold). As an exercise, you should check that every derivation $X_p$ at the point $p$ is given by $X_p(f)=\frac{d}{dt}\big\vert_{t=0}f(\gamma(t))$ for some smooth curve $\gamma$ such that $\gamma(0)=p$. This is why the two definitions are equivalent.

Alternatively, in local coordinates $(x^i)$, the derivations $\frac{\partial}{\partial x^i}\big\vert_p$ form a basis for $T_pM.$ It's common to write tangent vectors as a linear combination of these basis vectors.

 

[giulio_hep]

Ok so you only require $\gamma(0)=p$ and not that $\gamma'(0)$ is a pre-existing tangent vector at that point.

 

[zinq]

The nice thing about the definition of tangent vector based on derivations is that it does not depend on a choice of coordinates. (But a priori it is a bit unclear how a derivation connects with the geometry.)

(Ultimately one wants to know what kinds of things depend only on the differentiable structure of a manifold (or a submanifold of Euclidean space, like a sphere), and which things depend on additional structure like a Riemannian metric or a connection.)

 

[giulio_hep]

For me it's interesting to understand the physical interpretation of the cotangent space in classical mechanics first and then in quantization... Maybe my next question will be about the Maslov line and its cohomology class...