When we apply these concepts to physics, where do we put the UNITS in vector spaces and manifolds? Do units have a clear mathematical meaning?

Question

We know that the space of all displacements is a vector space.

The vector space is defined as a mathematical object $(V,k,+,\cdot)$ such that it satisfies the 8 properties, where $k$ is a field.

We notice that the field must not contain any unit, because it has to be closed under multiplication. for example, if we have a number "$5 \mathrm{m}$" in a field, then we must have the number "$25\mathrm{m}^2$" as well, which makes the space contain not only displacements.

Then the basis vectors of the space should contain the unit and the coordinates should contain only the scaling factors. The only way of writing coordinates that actually make sense should be like $$(1,2,3)_{(\mathrm{right}1\mathrm{m},\mathrm{front}1\mathrm{m},\mathrm{up}1\mathrm{m})}$$ instead of like

$$(1\mathrm{m},2\mathrm{m},3\mathrm{m})_{(\mathrm{right},\mathrm{front},\mathrm{up})}$$ while the latter seems to be what many people tend to write.

Now, we might stop caring that much about linearity. After all, as we

1. frequently use polar coordinates which are not linear anymore;
2. have generalized coordinates in theoretical mechanics that care even less about linearity;
3. have GR which states that the world is not linear at all.

2 and 3 both points to the concept of a manifold. and we consider that the coordinate is just a map from part of the manifold to a set of stuff, might it contain units or not. Here we can have our "$(r,\theta)$" coordinates (one has units and one has not) back.

But this has another problem, because we need the concept of a tangent space, and the tangent space should have a basis of $\frac{\partial}{\partial q^i}$. if $q^i$ have the unit of $\mathrm{m}$, then the basis should have $1/\mathrm{m}$. Yet in lagrangian mechanics, we need $\dot{q_i}$ which have unit $\mathrm{m}/\mathrm{s}$ live in the tangent space.

Also, the definition of manifold might ban units in the coordinates as well, since the definition requires a "homeomorphism to open subset of $\mathbb{R}^n$".

Should we just take the units away from all physical objects to make them work properly in math, or is there as way to make units function well in physics?

Answer 1

I think taking the formal tensor product of (say) $\mathbb{R}\otimes m$ for a length (meter) is good enough, in analogy how one can complexify vector spaces. Now use the natural convention of $(\mathbb{R}\otimes m) \times(\mathbb{R}\otimes m)$ (and universal property of tensor products) being $\mathbb{R}^2 \otimes m^2$.

This should be generalizable to arbitrary units.

Answer 2

A similar question was asked on Maths Stack Exchange, and the top answer was that dimensional object live in a tensor product of 1-dimensional graded vector spaces.

This builds on an essay by Terence Tao on the mathematical structure of dimensional analysis. He notes there are two ways of formalising it, a "parametric" approach

in which dimensionful objects are modeled as numbers (or vectors, matrices, etc.) depending on some base dimensional parameters (such as units of length, mass, and time, or perhaps a coordinate system for space or spacetime), and transforming according to some representation of a structure group that encodes the range of these parameters

and an "abstract" model

in which dimensionful objects now live in an abstract mathematical space (e.g. an abstract vector space), in which only a subset of the operations available to general-purpose number systems such as $\bf R$ or $\bf R^3$ are available, namely those operations which are “dimensionally consistent” or invariant (or more precisely, equivariant) with respect to the action of the underlying structure group.

The structure group can be the diffeomorphism group on a manifold and one gets GR-invariant dimensional analysis in both cases. I think there is still some real issues about how you define the connection so you can carry one dimensional object to another point, but the underlying mathematical structure of what dimensionful objects "are" seems straightforward.

Answer 3

It is good that you are already familiar with the concepts introduced by GR. Then I can just directly tell you that we are supposed to simply lay any coördinates (that work), and forget about trying to express positions as vectors. Instead, the coördinates can be used to set up a basis of tangent vectors by doing derivatives on those coördinate variables. General vectors are then linear combinations of these coördinate basis vectors. Proper time is a well-defined invariant scalar quantity suitable to use as a time parameter to enact differentiation, so that 4-velocity 4-vectors are well-defined. The linear combination of the 4-velocity 4-vector components are then defined to be contravariant quantities, as opposed to the covariant coördinate basis vectors, so that the 4-vectors themselves living in the abstract tangent spaces of the abstract manifold are well-defined (invariant). The changes in their components are only due to us changing coördinates describing them, not physically actually changing.

Now, if you are using polar coördinates, say $t,\varrho,\varphi,z$ of the standard cylindrical coördinates, then you are correct to note that the velocity component $v^\varphi$ clearly should not be carrying length units, at most only the reciprocal of time units.

However, the covariant coördinate basis vectors actually carry the units. You can check for yourself that if in flat Euclidean spacetime where we can express position as vectors, i.e. $\vec r=\begin{pmatrix}t,\varrho\cos\varphi,\varrho\sin\varphi,z\end{pmatrix}^T$ then all the coördinate basis vectors carry correct units; $\frac{\partial\vec r}{\partial\varrho}=\hat{\vec\varrho}$ has unit length but $\frac{\partial\vec r}{\partial\varphi}=\varrho\,\hat{\vec\varphi}$ will be longer the farther away from origin you are.

Most textbooks in the rest of physics does not want to teach these things properly, and so they simply use vierbeins, hence doing things like $\frac1\varrho\frac\partial{\partial\varphi}$ without properly explaining why we are doing them. In a sense, they are insisting upon manipulating their basis vectors to be of unit length, and by doing so, they get agreement with centuries old maths notation. Then they think they can get away without explaining whys and hows. Students are so confused that they do not have the correct vocabulary to articulate how confused they are. It is also because of these nuisances that the wonderful mathematical machinery that you learn whilst learning GR might sometimes not work for other parts of physics.

However, this kinda demystifies the confusion you have: For a tensorial quantity that has units, some of the units are carried in the components (e.g. 4-velocity's proper time derivative), and some units are actually carried in the covariant coördinate basis vectors. When we express some quantities in matrix form, sometimes we write down the units along with the components; those are hints, not formally there.

Now, of course, we rarely actually deal with the covariant coördinate basis vectors. This is no problem, because the metric, which is defined as the scalar product of the covariant coördinate basis vectors, is another good place to plug the units.

Note that I have not restricted my discussion above to use sensible, all-directions-using-same-units convention. If you want to use metres for vertical direction and nautical miles for horizontal directions, you are free to use that.