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Very elementary differential geometry questions

  1. Jan 18, 2012 #1
    I have decided to attempt to pick up some differential geometry on my own, and I am trying to get some traction on the subject which I do by trying to reduce it to familiar and simple cases.

    This is a trivial case, I know, but it will go a long way in advancing my understanding. Suppose the manifold of interest is the surface of a 2d-sphere (embedded in a 3d Euclidean space) and consider the tangent plane at point p. According to elementary differential geometry, a basis for this tangent space is [itex]\partial[/itex][itex]\mu[/itex].

    Now according to elementary linear algebra, one basis for this space would be:

    B= { [1 0], [0 1] }. These are written as row vectors because I don't know how to write them as columns here.

    My question: How do I get the B basis from the partial derivative basis from differential geometry?


    Another question that I have comes about when I get confused because the differential geometry formalism is introduced talking about a paramaterized curve in a manifold. What is the relationship between this paramaterized curve and the coordinate curves?
  2. jcsd
  3. Jan 21, 2012 #2
    The tangent space at given point is unique, but it can have many different bases. They are usualy denoted [itex]\frac{\partial}{\partial x^1}[/itex], [itex]\frac{\partial}{\partial x^2}[/itex]...

    Which basis is used doesn't depend on surrounding euclidean space, but on chosen coordinate chart. Coordinate chart gives you diffeomorphism between some subset of the manifold and ℝn, so when you for example want to differentiate real-valued function on manifold, you take it through coordinate chart to ℝn and do usual calculus there.

    When you take as manifold ℝn itself, I'm sure you can easily imagine usual partial derivative basis at any point of this manifold, maybe as arrows. Through coordinate chart you just push them to any other manifold.
  4. Jan 22, 2012 #3


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    To understand this, you need to understand the definition of the "tangent space at p", denoted by TpM if the manifold is denoted by M. The simplest (but probably least intuitive) way to define it is to say that it's the vector space spanned by the partial derivative operators associated with a coordinate system ##x:U\to\mathbb R^2## (where U is an open subset that contains p):

    Denote the manifold (in this case the sphere) by M. A function ##f:M\to\mathbb R## is said to be smooth if all partial derivatives up to arbitrary order of ##f\circ y^{-1}## exist for all coordinate systems y. For all p, all coordinate systems y whose domain includes p, and all i=1,2, define
    $$\frac{\partial}{\partial y^i}\bigg|_p f =(f\circ y^{-1})_{,i}(y(p)),$$
    for all smooth f. The notation ,i denotes partial differentiation with respect to the ith variable. Now we define TpM by choosing a specific coordinate system x and saying that TpM is the vector space spanned by the partial derivative operators associated with x. (Linear combinations of two partial derivative operators u,v are defined by (au+bv)f=a(uf)+b(vf)).

    This definition is ugly because it depends on a choice of coordinate system. But it can be proved that the result is independent of the coordinate system used. The definition can also be stated in a coordinate-independent way:

    Let S be the set of all smooth ##f:M\to\mathbb R##. Define TpM to be the set of linear functions v:S→ℝ such that v(fg)=v(f)g(p)+f(p)v(g) for all f,g in S. Define a vector space structure on TpM by (u+v)(f)=uf+vf and (av)f=a(vf). Now it's possible to show that given any coordinate system x, with a domain that includes p, the corresponding partial derivative operators form a basis for TpM. The proof can be found e.g. in Wald's GR book or Isham's book on differential geometry.

    The most intuitive definition actually assigns the term "tangent space at p" to a different vector space than this one. This doesn't really matter since there's an easily defined isomorphism between the two spaces. Here the idea is to define a tangent vector at p as an equivalence class of smooth curves through p. Two smooth curves are considered equivalent if they "are going in the same direction" at p. This idea and the associated definition of TpM is discussed in this post.

    A curve is a function ##C:[a,b]\to M##. If x is a coordinate system defined on a set that includes the range of C, then ##x\circ C:[a,b]\to\mathbb R^n## is a curve in ##\mathbb R^n##. I guess this would be the "coordinate curve". (I don't use those terms myself, and I'm not sure how other people use them. This is just what immediately comes to mind).
    Last edited: Jan 22, 2012
  5. Jan 22, 2012 #4
    Thanks very much for the replies.

    I'm going to follow up with a question which may be totally wrong, but I'm trying to get some intuition. Is there a function or set of functions to which [itex]\partial[/itex]i is applied to get the standard basis { [1 0], [0 1] }?

    I'll explain my motivation, and I apologize for my imprecise presentation (I am a physicist, not a mathematician): I know that if I have a point p on M and some open set U of M that is diffeomorphic (homeomorphic?) to ℝn together with an associated coordinate system, then I can perform calculations as if I were performing the calculations in Euclidean space. I want to make the connection in my mind between the differential geometric definition of vectors and "traditional" vectors in Euclidean space.

    I agree that the partial derivatives form a basis of the tangent space since I think that I can prove it using the techniques of ordinary linear algebra. I don't have a problem with that. Maybe my conceptual difficulty lies with the fact that I'm thinking of the wrong space. I don't know. I just think that the partial differential operators somehow should reduce to the standard basis that we use to do calculations. I suspect that the partial derivative operators operate on the coordinate system associated with U to produce the standard basis.

    I'm still left with the question of how does the single component [itex]\partial[/itex]i basis vector produce a two-component euclidean basis vector.
  6. Jan 22, 2012 #5
    First, a disclaimer: I'm reading only slightly more advanced stuff(cotangent bundles, tensors...) at this very moment, and the stuff you are asking I've learned only few weeks ago.

    Well, there are coordinate functions, usualy denoted (xi), which are real-valued functions defined on some open set U of a manifold. They are determined by a coordinate chart defined on U. For example, coordinate functions on R3 for identity function(i.e. one possible chart) are f(x, y, z) = x, f(x, y, z) = y etc. When you apply a tangent vector(with coordinates ai) to j-th coordinate function, you get [itex]a^i\frac{\partial}{\partial x^i}x^j = a^i\frac{\partial x^j}{\partial x^i} = a^iδ^j_i=a^j[/itex], so it picks j-th component of that tangent vector. There is also reverse process, given n smooth functions and a coordinate chart, you can assemble vector field from it.

    if it's topological manifold only homeomorphic, if smooth manifold diffeomorphic

    It's not that straightforward, for example, in ordinary calculus you can tell distance between two points using derivatives, integrals etc... but on smooth manifold you can't. You do stuff on smooth manifolds only such that doesn't change with diffeomorphisms(and distance, as you can easily imagine, changes, for example a sphere of double size). One thing you can do on SMs is integrate, since you integrate covector fields(or differential forms which are generalization of that) that change in a way that precisely cancels out any change of coordinates, diffeomorphisms etc... Imagine you paint lines on a rubber baloon and when you path-integrate you just count number of lines your path passes through.

    Also I know that while vector fields are rarely compatible with smooth maps(i.e. smooth function between two SMs), every covector field is compatible with smooth maps and can be pulled back from range to domain. So when you have path f: [a, b]→M and you have covector field on M, you pull it back to R and integrate it there using ordinary calculus.

    I started a thread on similar topic in november, you should check it out.

    The process of defining tangent vector takes about 5 pages in my book(Lee) and I think details of this process are not so important, but it's good to read through it and remember key fact(exact definition of tangent vector etc..). It does reduce to standart basis, but only using coordinate chart.

    You probably misunderstood your book. Don't forget about summation convention.
  7. Jan 22, 2012 #6


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    When ##\frac{\partial}{\partial x^i}\big|_p## acts on a function, the result is a number, not a vector. So, no, nothing like that is possible.

    I'm not sure what you're asking here. How to see that ##T_pM## is 2-dimensional when M is a sphere? That depends on which of the three definitions of ##T_pM## we use. If we use the first one, it's 2-dimensional by definition, so there's nothing to prove. If we use the second definition, you'd prove it using that trick that you can find in Wald or Isham. (It's pretty hard to think of). If we use the third definition, then I'd do it by proving that the space is isomorphic to the one produced by the second definition.

    If you're asking how members of a 2-dimensional vector space can be anything but 2×1 or 1×2 matrices, then you probably need to refresh your memory about some basic linear algebra. A vector space is said to be n-dimensional if it has a linearly independent subset with n members, and doesn't have one with n+1 members. It doesn't have anything to do with what the vectors look like. See posts #5 and #7 here for a nice example of this.

    Post #3 in this thread about the relationship between linear operators and matrices will also be useful. This is one of the most important details in all of linear algebra, so you should make sure you understand it perfectly. It will make it easier to understand tensors. (Ignore the stuff after the quote box).
    Last edited: Jan 22, 2012
  8. Jan 23, 2012 #7
    Thanks, you've both given me a lot to think about, and I really appreciate it.

    Fredrik, I'm going to follow your links and see if my confusion disappears. Of course, you're right that the definition of a basis has nothing to do with how a vector looks like. I still have a question, and if I still have it after reading your post again carefully and following your links, I will post it.

    Alesak, do you find Lee to be a good book? One of my biggest problems is finding a book that works for me ... books tend to be either so abstract as to be impenetrable for me or too simplistic (Geometrical Vectors by Weinreich). I'm not used to reading graduate texts in mathematics.
  9. Jan 23, 2012 #8
    Well, here on the forums I´ve seen only high praise for Lee, so I guess you can´t go wrong with it. For me it´s not an easy ride(I´m undergrad economist lol) but I doubt there is a book that will make learning it efortless.

    It´s quite detailed(IMHO good thing if you have time since SMs is important topic) and you need moderate knowledge of topology(but there is no need to go bonkers with it), good multivariable analysis and good linear algebra. Some knowledge of groups, modules... and categorical methods is useful. He says somewhere in introduction that his previous book on topological manifolds had been written as textbook on prerequisities for smooth manifolds, but I think there is way too more than strict prerequisites.

    But of course, it depends very much what background do you already have.
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