How can a set of components fail to make up a vector?

Question

Many books in Physics insist to define vectors are objects with components with the property that the components transform in a proper way under a change of coordinates. Now, in mathematics, on the other hand, vectors as geometrical objects (rather than the algebraic objects from linear algebra) belong to the realm of Differential Geometry. In that case, we have a smooth manifold $M$ and a point $a\in M$. A vector at $a$ can be defined:

1. As an equivalence class of smooth curves passing through $a$, which intuitively, at that point are going in the same direction.
2. As a point derivation, that is, a derivation on the algebra of germs of smooth functions at $a$.

This defines the tangent space $T_a M$. This is a real vector space with dimension $n = \dim M$, and hence $T_a M \simeq \mathbb{R}^{n}$. In particular, this means there is a bijection between $T_a M$ and $\mathbb{R}^n$ so that given any tuple of components, they do form a vector on $T_a M$. This seems to be against the physicists' definition since there's nothing imposed on the components to form a vector.

On the other hand, we can put all tangent spaces together to form the tangent bundle $TM$. We define then vector fields as sections of that bundle, that is, mappings $X: M\to TM$ such that $\pi\circ X = \operatorname{id}_M$ where $\pi : TM\to M$ is the natural projection. $X$ should also be continuous and differentiable, and of course, it also has components given by

$$X = X^i \dfrac{\partial}{\partial x^i}$$

Now, given a set of component functions $X^1,\dots, X^n$ I can't see how they fail to make up a good vector field. If the functions are differentiable, continuous, and if they obey the property that $\pi \circ X= \operatorname{id}_{M}$ then we are good to go.

So my questions are:

1. What is really the point in giving so much emphasis on the way vectors transform, coming to the point that we use this property to even define vectors?

2. When physicists talk about defining a vector using transformation properties, are they really talking about vector fields and changes of coordinates on a manifold or vectors and changes of basis on each tangent space?

3. How can a set of components (or component functions) fail to make up a vector (or vector field)?

Answer 1

Vectors can be defined in multiple different ways. Here I will show the four ways and explain how they are equivalent$^1$.

1. An equivalence class of curves. Given a curve parameter $t$, curves are considered equivalent if they have equal zeroth- and first-order derivatives at $t=0$. In other words, a vector at a point $p$ is an equivalence class $[\gamma]$ such that $$x^i(\gamma(t))=x^i(p)+tv^i+O(t^2),\quad v^i:=\left.\frac{d}{dt}\right|_{0}x^i(\gamma(t))$$ Here the basis vectors are defined as the equivalence classes $[\gamma_i]$ where $\gamma_i$ is the curve of constant $x^i$.

2. A linear functional on the algebra of the algebra of (smooth) functions$^2$ that is a derivation. This definition of the vector is the directional derivative one. Given a curve $\gamma$, we define the tangent vector $v$ such that $$vf=\left.\frac{d}{dt}\right|_{0} f(\gamma(t))$$ The basis vectors are just the directional derivatives along curves of constant $x^i$. We also see that this definition of the vector is compatible with the equivalence class definition, since the derivative does not depend on $\gamma\in[\gamma]$.

3. A first-order differential operator. Here we write$^{3,4}$ $$v=v^i\partial_i|_p,\quad\partial_i:=\frac{\partial}{\partial x^i}$$ To see how this relates to the second definition, apply this to a function: $$vf=v^i\partial_i|_p f$$ Now use the chain rule in the second definition to obtain $$v^i=\frac{dx^i}{dt}$$ This also makes a connection to the definition of $v^i$ in the first definition. We note further that partial derivatives, under a change of coordinates $x\rightarrow x'$ transform with the inverse of the Jacobian. For $v$ to then be coordinate invariant, $\{v^i\}$ must transform with the Jacobian.

4. An $n$-tuple of (real) numbers that transforms with the Jacobian. All we say here is that $v^i$ is a vector if $v^i\rightarrow J^i{}_j(p)v^j$. This transformation rule is consistent with that of definition 3.

Armed with this knowledge, let us tackle your questions.

1. It is very easy to grasp at the introductory level. It also generalizes nicely to vectors in different contexts. For instance, we can have vectors under rotation, where the "Jacobian" is simply a rotation matrix. Another example is a vector in $SU(N)$,$^5$ which changes with a unitary matrix. In other words, a vector is something that transforms like a vector. This then generalizes to tensors. A tensor is something that transforms like a tensor.

2. A vector field is just an object that maps a vector to each point on the manifold. Vectors and vector fields transform the same way, the only difference is that the coefficients in the linear expansion are nonconstant and the partial derivative basis vectors are not restricted to a point.

3. The tuple $\{x^i\}$ is not a vector. It does not transform properly. Some other examples can be found in general relativity. The acceleration vector $\ddot x^i$ is not a vector, despite being a tuple. This is because it does not transform properly. We also have the quantity appearing in the geodesic equation, $\Gamma^\lambda{}_{\mu\nu}\dot x^\mu\dot x^\nu$. The Christoffel symbols famously do not transform as tensors, so this combination is not a vector.

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$^1$ See [1] p. 23 ff or [2] p. 589 ff for more details.

$^2$ This post makes no distinction between functions and germs.

$^3$ We don't have to use partial derivatives, but the partial derivatives form a very convenient basis of the tangent space. See [3] p. 15 for a simple proof.

$^4$ Note that the partial derivatives are to be evaluated at the point $p$.

$^5$ Also known as the fundamental representation.

References:

[1] M. Fecko, Differential Geometry and Lie Groups for Physicists (2006).

[2] N. Straumann, General Relativity (2013).

[3] R. Wald, General Relativity (1984).

Answer 2

When physicists say that a vector is an n-tuple that transforms according to... they expect you to guess a lot that isn't said. What they mean is that for each basis you are given an n-tuple of numbers. And when you take the matrix that gives you the change of bases for any two bases and you apply the formula in their "definition" the first n-tuple goes to the second an so on. Hence these n-tuples are the coordinates of a vector with respect to each basis.

edit: In his answer 0celo7 gives the four ways to define tangent vectors at a point on a manifold. So here is another way for those that are algebraist in heart. Let $M$ be a manifold and $p$ a point on it. Denote by $\mathcal O_p$ the ring of germs of smooth functions at the point. It is a local ring in the sense that it has a unique maximal ideal $\mathcal m_p$. The quotient $\mathcal O_P/\mathcal m_p$ is isomorphic to $\mathbb R$ and $\mathcal m_p/\mathcal m^2_p$ is naturally a vector space over $\mathcal O_P/\mathcal m_p$ i.e. it is a real vector space. That is called the cotangent space of the manifold at the given point. Its dual $\left(\mathcal m_p/\mathcal m^2_p\right)^*$ is the tangent space. One can check that this is equivalent to any of the four definition.

Answer 3

OP wrote (v3):

Now, given a set of component functions $X^1,\dots, X^n$ I can't see how they fail to make up a good vector field. If the functions are differentiable, continuous, and if they obey the property that $\pi \circ X= \operatorname{id}_{M}$ then we are good to go.

It is (implicitly) implied by OP's notation that

1. the component functions $X^1,\dots, X^n\in C^{\infty}(M)$ are globally defined functions.

2. the coordinate functions $x^1,\dots, x^n\in C^{\infty}(M)$ are globally defined functions.

However, there are many examples of differentiable manifolds $M$, that don't have a global coordinate chart, e.g. the 2-sphere $S^2$.

The general notion of a vector field should not rely on whether there exists a global coordinate chart, or not.

Answer 4

I think that the physicists and mathematicians in this situation take two different ways to obtain the same thing. As Borges would have said, everyone is either platonist or aristotelian; in this case (and probably always) the mathematicians being the former and the physicists the latter.

From the mathematical point of view, you the general structure (category) of manifolds and tangent spaces and so on, of which the physical manifolds of classical and relativistic physics are just special cases (keep in mind that the manifolds of relativity are locally homeomorphic, as metric spaces, to the minkowski space and not the euclidean one: the set is always $\mathbb{R}^n$, but the metric is different).

From the physical point of view, you have objects that you can observe in the real world, and that obey to certain rules, most notably (in this context) they transform in a peculiar way under change of coordinates between different reference frames. From these rules it is useful to infer a general structure, in order to make interesting predictions. It turns out that the suitable metric structure of the space-time "accidentally" fits in the definition of riemannian manifolds (locally minkowski/euclidean).

Within this new powerful structure, the galilean/lorentz transformations of coordinates are just particular maps between charts in the manifold, that can be seen as the local version of endomorphisms of the riemannian manifold (transformations that associate to each point of the manifold another point of the same and that preserve the riemannian structure of the manifold itself, most notably its metric).

Also, on the other hand, given a manifold it is possible to characterize its endomorphisms, and therefore recover the physical transformations as a particular case.

The rest is in some sense a consequence: given an endomorphism of the manifold, this induces an endomorphism of the tangent bundle and thus you will have how vector fields transform under the transformation, and so on. But this is fixed, as the concept of vectors, forms etc. once you have given the manifold.

Answer 5

Take Cartesian coordinates for the real plane and transform them into polar. Does the set of coordinates $(x,y)$ transform as a vector? If you work out this example you'll see that this transformation, unlike linear ones, doesn't involve the Jacobian of the transformation. In the linear case this is just an accident.

Answer 6

There's been some good discussions linked and 0celo7 wrote the answer with great brevity, but I'll take a crack at adding some exposition. Let's consider situations of rotation:

We can consider clockwise rotation (for our cartesian basis $\{e_i\}$) for some vector $V$,

$ \hat x' = \hat x\cos(\theta)+\hat y\sin(\theta) $
$ \hat y' = -\hat x\sin(\theta)+\hat y\cos(\theta) $

Embedded in this, we see the matrix $R(\theta)$. If we're doing a true coordinate transformation, we must also consider how our components transform as well,

$V = v^ie_i = v'^ie'_i = (R^{-1})^i{}_jv^jR^k{}_ie_k $

What we need to be careful to recognize is that we've already defined our vector based on its transformation. How about we consider a different transformation $R'$,

$ \hat x' = \hat x\cos(\theta)+\hat y\sin(\theta) $
$ \hat y' = +\hat x\sin(\theta)+\hat y\cos(\theta) $

If you then explore this new "vector" object $W$ that transforms with $R'$, you'll find that your "vector" wildly wobbles about as you change your coordinates. We've elected to not call such objects vectors in Cartesian space. There may some other space this transformation and associated objects call home, but it isn't this manifold. For similar reasons, it's nonsensical to apply Lorentz transformations to 4D space with a $(+,-,-,+) $ signature.

In short, for every differential geometry, there exists transformations which define the objects in that geometry which remain unaffected by your arbitrary perspective--essentially embodying whatever symmetry that geometry contains.

Answer 7

From a strictly mathematical point of view, the reason why the notion of:

$$\text{“$n$ globally defined scalar fields $f_i\colon M \to \mathbb R $”}$$ is different from the one of: $$\text{“a vector field $X\colon M \to TM$”}$$ is indicated in Qmechanic's answer. The notion of vector field is intrinsic, that is, it doesn't depend from the choice of a chart on your manifold. When you have a set of smooth functions, you have to explicitly choose a chart to make up a vector field from them, and there's no canonical way to do this.

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Citing Sakurai:

The operator is different from the representation of the operator just as the actress is different from a poster of the actress.

It's more or less the same thing.