I've been reading the book Gauge, Fields, Knots and Gravity by Baez.

A tangent vector at $p \in M$ is defined as function $V$ from $C^{\infty}(M) $ to $\mathbb R$ satisfying the following properties:

  1. $V(f+g)=V(f) + V(g)$.

  2. $V(\alpha f)= \alpha V(F)$.

  3. $V(fg) = V(f)g(p) + V(g)f(p)$.

Can someone explain me what is the physical interpretation of tangent vectors and the above definition?

  • 3
    $\begingroup$ This one often causes bewilderment. The advantage of defining tangent vectors this way is that it is very easy to state without introducing local coordinates. As Zhen Lin points out, this definition corresponds to the geometric one by thinking of directional derivatives, i.e. the action of a tangent vector on a function is the directional derivative of the function in that direction. $\endgroup$ – Cheerful Parsnip Oct 24 '11 at 20:43

Perhaps you would prefer a different definition: "A tangent vector is a possible time derivative at $t=0$ of a smooth curve $\gamma:\mathbb R\to M$ such that $\gamma(0)=p$".

That's rather more intuitive. Unfortunately it is not really convenient as a definition. First, we need to figure out when two tangent vectors at $p$, represented by $\gamma_1$ and $\gamma_2$ are the same. Well, easy-peasy, we just select a chart that covers $p$ and use that to express $\gamma_1$ and $\gamma_2$ as $\mathbb R\to\mathbb R^d$. Then we ask whether $\frac{d\gamma_1}{dt}=\frac{d\gamma_2}{dt}$, and if so the vectors are the same. Then, however, we need to do a lot of nitty-gritty work proving that it doesn't matter which chart we use, and so forth.

Second, we'd like the tangent spaces at $p$ to be a vector space, so we need to define addition of tangents. And it's not really clear how to do that with velocities, so again we have to select a chart, derive the two curves, and find a new curve whose derivative is the sum of the derivatives. Again, lots of gritty-work showing that it doesn't matter which chart we use. And so forth.

All in all, this is cumbersome, and it is also unaesthetic because we don't like to have definitions where it is not obvious that they define something in the first place, and because we'd like to do as much of differential geometry as possible without referring to specific charts all of the time.

Therefore, we use a trick. Instead of using the curves $\gamma$ directly as a definition of tangents, we say: For an arbitrary (differentiable) scalar field $f$, look at the composition $t\mapsto f(\gamma(t))$ and take its derivative at $0$. It turns out that the mappings $f\mapsto \frac{d}{dt}f(\gamma_1(t))$ and $f\mapsto \frac{d}{dt}f(\gamma_2(t))$ are the same (for all $f$) if and only if $\gamma_1$ and $\gamma_2$ represent the same vector according to the clumsy definition. Therefore we can use this map to represent the tangent vector, and then we don't need to quotient with any iffy equivalence relation. And better yet, linear functionals from scalar fields to $\mathbb R$ are naturally a vector space, with a vector-space structure that turns out to agree with the clumsy but intuitive chart-based definition.

Thus, formally a tangent vector is just a scalar operator on the space of scalar fields, satisfying certain axioms that guarantee it behaves like a differential operator should. Informally you should probably think of it as a velocity through the point in question. Its operator incarnation measures how quickly a scalar field changes, seen from a moving point whose instantaneous velocity is the vector.

Alternatively, you can think of a tangent vector as something that combines with a gradient to form a scalar in a bilinear manner. Its formal incarnation is again a bit strange under this interpretation, because it comes with a built-in gradient operation and cannot be applied to a gradient you've already prepared. But that's just for technical reasons.


What this is defining is a derivation $C^\infty(M) \to \mathbb{R}$. The intuition is that such a $V$ is a directional derivative of some kind: directional derivatives are linear and satisfies the Leibniz law, and that's exactly what your axioms are stipulating. Obviously, a directional derivative defines a direction and vice versa, so this is one way of defining a tangent vector at a point.

Another equivalent approach is to regard a tangent vector at a point as an equivalence class of curves through a point. Here we take a more hands-on approach and say that because we know how to define curves on a manifold, we can exploit our knowledge of one-variable calculus to differentiate along a curve; but again, following our intuition that two curves can be, well, tangent at a point, we need to take equivalence classes in order to get a one-to-one correspondence with tangent vectors.


There are a lot of good answers to this already, but since I just went through this material recently, I'll offer my perspective. The key fact that made viewing "tangent vectors" as derivations make sense to me is that there exists an isomorphism between all derivations at a point $p$ and all vectors based at $p$. Specifically, if

  1. $X$ is an open subset of $\mathbb{R}^n$ and $p \in X$
  2. $\mathcal{D}_p(X)$ represents the vector space of all derivations at $p$
  3. $T_p(X)$ represents the vector space of all vectors $v$ based at $p$

we can define a linear bijection

$$ \phi:T_p(X) \rightarrow \mathcal{D}_p(X) $$


$$ \phi(v) \rightarrow D_v $$

where $D_v$ indicates the directional derivative operator in the direction $v$. Note that this definition makes sense because $D_v$ is itself a derivation.

Showing that $\phi$ is linear and injective is straightforward; proving surjectivity takes a little more work and requires some tools from calculus; specifically Taylor's theorem with remainder. Tu's Introduction to Manifolds works all of this out in detail on pages 13-14.


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