Carroll writes in his Spacetime and Geometry book on page 68 that

"[...] in fact, however, we could just as well have begun with an intrinsic definition of one-forms and used that to define vectors as the dual space. Roughly speaking, the space of one-forms at $p$ is equivalent to the space of all functions that vanish at $p$ and have the same second partial derivatives. In fact, doing it that way is more fundamental, if anything, since we can provide intrinsic definitions of all $q$-forms (totally antisymmetric tensors with $q$ lower indices), etc."

Could somebody explain this equivalency in a bit more detail?

  • $\begingroup$ What is p element of? A vector space or an abstract point set? $\endgroup$
    – lalala
    Sep 16 '19 at 14:53
  • $\begingroup$ Assuming $p$ is a point in the tangent plane to the manifold $M$, then a function $f_{p}$ is $0$-form, $df_{p}$ is a $1$-form, and $ddf_{p}=0$ (or $d^2f_{p}=0$) - where $d$ is the exterior derivative and $d^{2}=0$ is the closure property. And not to go off the deep end, but the space of differential forms is determined by the topological properties of the manifold. See De Rham cohomology. $\endgroup$ Sep 16 '19 at 21:48

If $f\in C(M)$ is a smooth function on $M$ (I am dropping the "$\infty$" sign, since all functions shall be smooth), and $v\in T_xM$ is a tangent vector then the assignment $v\mapsto v(f)$ is a linear operation $T_xM\rightarrow\mathbb R$, so in some sense, a smooth function is also a linear functional on $T_xM$, thus is an element of the dual space $T^\ast_x M$.

There is a problem with this interpretation however, is that dual spaces are required to be separating. Eg., if $V$ is a vector space, and $V^\ast$ is its (algebraic) dual, we demand that for any $\omega,\omega^\prime\in V^\ast$ can be told apart by how they act on elements of $V$. In other words, if $\omega(v)=\omega^\prime(v)$ for all $v\in V$, then we must have $\omega=\omega^\prime$.

There are however functions $f,f^\prime\in C(M)$ (prime is not a derivative here) such that $v(f)=v(f^\prime)$ for all $v\in T_xM$. The prudent thing to do then is to identify the annihilator $T^0_xM\subset C(M)$ consisting of functions $f$ that satisfy $v(f)=0$ for all $v\in T_xM$. Then we clearly have $T^\ast_xM=C(M)/T^0_xM$.

This procedure is best carried out in two steps. First one should note that all tangent vectors have the following two properties:

  • If $v\in T_xM$ is an arbitrary vector and $f\equiv c$ is a constant function, then $v(f)=0$.
  • If $v\in T_xM$ is an arbitrary vector and $f=gh$, where $g(x)=h(x)=0$ (but only necessarily for the specific point $x$, they do not have to vanish identically), then $v(f)=0$.

One may then define the space $\mathcal I_x\subset C(M)=\{f\in C(M):\ f(x)=0\}$ consisting of functions that vanish at $x$. This a subring of $C(M)$ and is also a two-sided ideal, since it is stable under multipliction from outside of $\mathcal I_x$.

We have for any $v\in T_xM$ the relation $v(f)=v(f-f(x))$, since $f(x)$ is just a constant (the value of $f$ at $x$), and the map $f\mapsto f-f(x)$ is a projection from $C(M)$ to $\mathcal I_x$, so the action of $T_xM$ on $C(M)$ descends to an action on $\mathcal I_x$ without any loss of information. Thus, we succeeded in "quotienting out" a portion of the annihilator $T^0_xM\subset C(M)$ without actually taking a quotient, since the only constant function remaining in $\mathcal I_x$ is the one that is identically zero.

There is still the issue of an arbitrary tangent vector $v$ being zero on products of the form $f=gh$ with $g,h\in\mathcal I_x$. We first note that ideals can be multiplied, eg. $\mathcal I_x^2$ is the ideal in $C(M)$ generated by expressions of the form $fg$ where $f,g\in\mathcal I_x$. The "second" property of tangent vectors corresponds precisely to the statement $$\text{All tangent vectors annihilate }\mathcal I_x^2.$$

Thus we can get rid of this subspace by taking the quotient $\mathcal I_x/\mathcal I_x^2$. The action of an arbitrary $v\in T_xM$ tangent vector factors through the quotient. In fact, let $[\cdot]$ denote the quotient projection $\mathcal I_x\rightarrow \mathcal I_x/\mathcal I_x^2$, then we define the differential of a function $f\in C(M)$ at $x\in M$ as $$ \mathrm df_x=[f-f(x)]\in\mathcal I_x/\mathcal I_x^2, $$ then if we define $$ \mathrm df_x(v)=v(f),$$ this doesn't depend on which $f$ we choose from the same class.

At this point, we cannot be sure that we cannot quotient out any more subspaces from $C(M)$, however one can prove that the space $\mathcal I_x/\mathcal I_x^2$ is finite $n$ dimensional (and it is only a linear space, the algebra structure is lost during the quotient, since products of functions are quotiented into the zero element).

The standard proof involves a Taylor expansion with remainder. Heuristically, without using the precise form of the remainder term, we can write $$ f=f(x)+\partial_\mu f(x)(x^\mu-x^\mu(x))+\mathcal I_x^2, $$ where the last expression means that the remainder is of the form $(x^\mu-x^\mu(x))(x^\nu-x^\nu(x))g$, so it is in $\mathcal I_x^2$. Applying the differential $\mathrm d_x$ into this expression gives $$ \mathrm df_x=\mathrm d(f(x))_x+\partial_\mu f(x)\mathrm d(x^\mu-x^\mu(x))_x+\mathrm d(\mathcal I_x^2)_x=\partial_\mu f(x)\mathrm dx^\mu_x, $$ and since $\mathrm d_x$ is the composition of two projections it is necessarily surjective. Thus we see that for any local chart, the elements $\mathrm dx^\mu_x$ span $\mathcal I_x/\mathcal I_x^2$, and linear independence can also be easily established, thus $\dim T_xM=\dim(\mathcal I_x/\mathcal I_x^2)$. This however implies that $\mathcal I_x/\mathcal I_x^2$ is the dual space of $T_xM$, since there are no subspaces left to quotient out anymore.


The procedure $$ C(M)\rightarrow\mathcal I_x\rightarrow \mathcal I_x/\mathcal I_x^2$$ involves gradually removing all elements from the space of functions $C(M)$ which indiscriminately annihilate all tangent vectors, thus in this way we arrive at the "true" dual space of $T_xM$ where each element can be told apart by how they act on tangent vectors (or how tangent vectors act on them).

If wer use a generalized Taylor expansion in a chart $$ f=f(x)+\partial_\mu f(x)(x^\mu-x^\mu(x))+\frac{1}{2}\partial_\mu\partial_\nu f(x)(x^\mu-x^\mu(x))(x^\nu-x^\nu(x))+O((x-x(x))^3), $$ we see that the first, constant term $f(x)$ gets cut off when we move from $C(M)$ to $\mathcal I_x$, and we also see that the secomnd, third and all higher order terms contain factors of the form $(x-x(x))^2$, thus they belong to $\mathcal I_x^2$. Taking the quotient $\mathcal I_x/\mathcal I_x^2$ essentially involves getting rid of every term in the Taylor expansion aside from the first order term, and what we remain with is $ \partial_\mu f(x)(x^\mu-x^\mu(x))$.

Now, we know that covectors on a manifold are essentially first-order partial derivatives of functions. What we did by this quotienting procedure (by moving $f$ to $\mathrm df_x$) was essentially that we differentiated the function $f$ in a purely algebraic manner. Thus, we constructed the space of first-order partial derivatives of functions, eg. the cotangent space.


EDIT: I agree with Kostya that there is likely a typo in Carroll's book, and that the equality of first (not second) derivatives is what's required. I haven't thought about this for long, though, so it's possible I'm mistaken here.

Let the set of all smooth functions from a manifold $M$ to $\mathbb R$ be denoted $C^\infty(M)$. As you know, a tangent vector $\mathbf X_p$ at a point $p\in M$ is a linear map which eats functions $f\in C^\infty(M)$ and spits out real numbers, interpreted as their directional derivatives along $\mathbf X_p$.

In a particular coordinate chart with coordinates $y^a$, we can associate $\mathbf X_p$ to the differential operator $$\mathbf X_p = X^a\frac{\partial}{\partial y^a}$$ where $X^a$ are called the components of $\mathbf X_p$ (in the chart). The action of $\mathbf X_p$ on a function $f$ is then

$$\mathbf X_p (f) = X^a \frac{\partial f}{\partial y^a}$$

The set of $1-$forms $T_p^*M$ is the set of linear maps from $T_pM$ to $\mathbb R$; in other words, the linear maps which eat vectors and spit out real numbers. Given some smooth function $f\in C^\infty(M)$, we can define a one form $df:T_pM \rightarrow \mathbb R$ as follows:

$$df(\mathbf X_p) := \mathbf X_p(f)$$

This might seem trivial, but the idea is that we can construct one forms from smooth functions. This leads us to the question of whether $C^\infty(M)$ and $T^*_pM$ are essentially the same space wearing different hats.

The answer to that is no, not quite, because $C^\infty(M)$ is too big. Working in an arbitrary coordinate system, we can equate a one-form $df$ with its components

$$df = \left(\frac{\partial f}{\partial y^a}\right)_p dy^a$$

Clearly two functions which differ by an additive constant would produce the same directional derivative. We can solve this by shifting our focus from $C^\infty(M)$ to $C^\infty_0(M)$ - the set of smooth functions $f$ such that $f(p)=0$.

$C^\infty_0(M)$ is still too big, because two different smooth functions can have the same first derivatives at $p$. This leads us to the definition of the following equivalence relation.

Define the equivalence relation $\sim$ on the set $C^\infty_0(M)$ such that $f\sim g$ if $\frac{\partial f}{\partial y^a} = \frac{\partial g}{\partial y^a}$. This carves up $C^\infty_0(M)$ into equivalence classes of smooth functions which all share the same first derivatives at $p$. The set of all such equivalence classes is called the quotient set $Q\equiv C^\infty_0(M) / \sim$, and it is this set which is isomorphic to the set of one forms at $p$.

To see this, simply note that an element of $Q$ can be uniquely specified by a list of numbers which correspond to its partial derivatives at $p$ (note that this is not true of $C^\infty_0(M)$), and that a one-form can be uniquely specified by its components. These lists can then be put into one-to-one correspondence.

  • $\begingroup$ While all you wrote is true, where does the condituon that the second partial derivates agree come in? $\endgroup$
    – lalala
    Sep 16 '19 at 18:08
  • $\begingroup$ @lalala I agree with Kostya that there is likely a typo there. It is not listed among the errata for the book, though, so I may be crazy. $\endgroup$
    – J. Murray
    Sep 16 '19 at 19:07
  • $\begingroup$ @Kostya After having received this and the answers below I felt confused and asked Prof. Carroll directly the same question. His answer was: "Nope, it is correct. If the first derivatives were the same, they'd be the same one-forms." Now I am even more confused but did not bother him with further questions. $\endgroup$
    – hyportnex
    Apr 14 at 14:03

Could somebody explain this equivalency in a bit more detail?

The fundamental mathematical notion here is the idea of "equivalence class". If you have a set of things (in your case it is the set of functions on a manifold) and you have some equivalence relation between these things (in your case two functions are equivalent if they vanish at a given point and their derivatives are the same). Then your set will be broken into a number of non-overlapping subsets - the elements in each subset are all equivalent to each other. The conceptual jump is then to forget about the fact that each equivalence class is a (sub)set and just manipulate them as individual elements -- e.g. in your case, you can define addition/subtraction/constant multiplication on the equivalence classes of your functions.

What your quote says is that, mathematically, it is very convenient to define 1-forms as equivalence classes of functions at a given point. Although, it seems like this is a mistake:

...have the same second partial derivatives...

I think it should be "first" derivatives instead -- check this definition on Wikipedia.


The tangent and cotangent spaces to a finite-dimensional manifold at a point $p$ are finite dimensional vector spaces and hence canonically isomorphic to their own double duals. So you can either a) start with any of several equivalent definitions of the tangent space and then define the cotangent space to be the dual of the tangent space, or b) start with any of several equivalent definitions of the cotangent space and then define the tangent space to be the dual of the cotangent space --- and up to canonical isomorphism, it doesn't matter whether you follow strategy a) or strategy b).

The quote suggests that it's often most natural to follow strategy b), using this defintion of the cotangent space: First define a germ at $p$ to be an equivalence class of differentiable functions under the equivalence relation that sets two functions to be equivalent if they agree in some neighborhood of $p$. Then let $M$ be the set of all germs represented by functions that vanish at $p$, let $M^2$ be the subset of $M$ spanned by germs represented by products $fg$ with $f,g$ both in $M$, and define the cotangent space to be the vector space $M/M^2$.


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