Do you really mean R^n?

In mathematics and physics it is common to talk about \mathbb{R}^n when really we mean something else that can be represented by \mathbb{R}^n .

Consider mechanics or geometry, these are often represented as theories in \mathbb{R}^n, but really they don’t occur in a vector space at all! Look around you, a three-dimensional description of space probably seems reasonable, but where’s the origin? [Perhaps the centre of your eyes could be an origin, but someone else would disagree with you]. Classical mechanics, special relativity and geometry are much better described as an affine space – which is a vector space without an origin.

Of course to do any calculations we need to choose an origin in the affine space and choose some axes, and this allows us to repesent our space as \mathbb{R}^n. But if we calculate the volume of a box (relative to another box) using two different origins we expect to get the same answer. This is precisely what the affine structure says: our results are independent of our choice of origin (that is they are invariant/covariant with respect to translations).

Now suppose you have a genuine n-dimensional vector space (perhaps you’ve chosen an origin in your geometry), we can now call this \mathbb{R}^n right? Wrong; \mathbb{R}^n has a natural ordered basis e_1=(1,0,\ldots,0), \ldots, e_n=(0,\ldots,0,1). Again, taking you as the origin in the space around you – what are the three basis vectors? There are lots of choices!

Alright so let’s choose n (ordered) linearly independent basis vectors, by identifying the the ith one with e_i (for the space around us we could choose (magnetic) North, (magnetic) East and (graviational) up), and this gives us axes. But there’s nothing super-special about these choices – again our box-volume should be independent of our choice of axes. This is encoded in the vector space structure: there’s no canonical choice of basis, and all physical quantities should be properly covariant with an appropriate change of basis.

Fine – we’ve taken an n-dimensional vector space and chosen an ordered basis – surely we can now call it \mathbb{R}^n? Well it depends on how much extra structure you have on your vector space. The ordered basis for \mathbb{R}^n gives it a canonical orientation. There is a canonical inner product on \mathbb{R}^n, the dot product \langle e_i, e_j \rangle = \delta_{ij}. (It is exceedingly rare to see the distinction made between \mathbb{R}^n as a vector space, an oriented vector space and an inner product space.) So your vector space must at least have an orientation and an inner product (and your choice of ordered basis must be oriented and orthonormal) – but this is a lot of extra structure to throw in.

A lot can be said for the geometry of affine spaces over a vector space (that is without a metric or orientation). There is no concept of absolute volume (this is given by a metric), but there is a relative notion of oriented volume for n-dimensional parallelograms (which is essentially “how many copies of parallelogram 1 fit into parallelogram 2” with the conditions that volume is translation invariant and scaling a side by some factor scales the volume by the same factor). In fact we can talk about the relative volume for any hyper-polygon (by cutting it into triangular pieces – a parallelogram is just two triangles stuck together) and approximate the relative volume of other objects such as spheres (this is how a lot of Greek geometry was done). There is also the notion of an affine transformation, one that preserves the affine structure (essentially a shift followed by a linear transformation), and a determinant which is the amount by which an affine transformation scales the volume of n-dimensional parallelograms.

If we push even further to locally affine transformations the determinant will naturally generalize to the Jacobian of the transformation, and one can define the exterior derivative. Or thinking of spaces that are locally affine leads to manifolds. But this is a whole essay in itself.

Of course we do often have more structure – Euclidean geometry, classical mechanics, Electrodynamics and special relativity all have metrics – a notion of absolute volume. But it is interesting to separate the metric independent behaviour (for instance in Electrodynamics the Maxwell equations \rm{d} \bold{F}=0) from the metric behaviour (\rm{d} \star \bold{F} = \bold{J}). Certainly it is useful to write equations in a form where they are explicitly independent of the choice of origin and basis (as opposed to writing them in \mathbb{R}^n) – this is the advantage of Lagrange’s method of mechanics over Newton’s.

We should explicitly distinguish these structures from $\mathbb{R}^n$. A vector space should just be denoted by a letter like V, with possibly a superscript to denote its dimension like V^{(n)}. An affine space could be denoted A^{(n)}, an affine space with a positive definite inner product is called Euclidean space E^{(n)} and an affine space with an inner product signature (n-1,1) is called Minkowski n-space M^{(n)}. I would advocate using E^{(m,n)} for an affine space with an inner product signature (m,n).

As I’ve said at the end of the day to do calculations we will go into a specific representation which may look like \mathbb{R}^n and this is fine – but it is important to state and develop the theory in ways that are independent of these choices.

End Note: Given a basis for a vector space V \{e_1,\ldots,e_n\} we can define a dual basis for V* \{e^1,\ldots,e^n\} by e^i(e_j) = \delta^i_j. The mapping of a basis onto its dual gives an isomorphism from V to V*, but this isomorphism isn’t natural.  This has a specific categorical meaning but roughly its because if we have a linear transformation L on V, L has a natural action on V*. In terms of the matrix of the transformation the action on V* is the adjoint matrix (that is the transpose or Hermitian conjugate depending if the matrix is real or complex). But the dual basis transforms by the inverse to maintain e^i(e_j) = \delta^i_j. So the adjoint action preserves the identification of bases (and so the isomorphism is independent of our choice of basis) if and only if T^* = T^{-1}, which means the matrix is orthogonal or unitary. However apriori there is no reason to only use this type of linear transformation so the identification is not canonical. If, however, there is a metric on our vector space we should only use linear transformations that preserve this metric, and in terms of an orthonormal basis, this means we only use orthogonal/unitary transformations. So a metric induces a canonical isomorphism between a vector space and its dual! It is not too hard to see conversely that an isomorphism between a vector space and its dual gives a metric.

This is another reason (related to the existence of a natural inner product) a vector space should not be represented by \mathbb{R}^n – it is canonically identified with its dual.

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7 Responses to Do you really mean R^n?

  1. Pingback: Do you really mean S^1? | physjam

  2. Stephen says:

    I’m not sure on this one, but how is the inner product defined given an affine space? The only place I’ve seen an inner product is in relation to a vector space.

    • physjam says:

      Good question.

      One way to put an inner product on an affine space is to choose an origin, and put a metric on that vector space, and insist that the metric is invariant under translations (that is if we choose a different origin we get the same metric).

      A more invariant way would be to introduce the notion of length. To every line segment there is associated a positive number we call the length, that satisfies some properties:
      i) It is invariant under translations.
      ii) It is additive over parallel lines.
      iii) It satisfies the Paralelogram law – this allows us to define angles by the polarization identity.

      Then a Euclidean space is an affine space with a notion of length, and a Euclidean transformation is a transformation that preserves length (i.e. a composition of translations and rotations). [We can define a sphere radius r and centre C as the set of all points distance r from C. Another way of thinking of Euclidean transformations is it is the set of all linear transformations sending spheres radius r to spheres radius r.]

      There’s a similar relationship between manifolds and Riemannian manifolds. You can make a Riemannian manifold by putting a metric on every tangent space of the manifold that varies nicely over the manifold, or you can consider a Riemannian manifold to be a manifold with some concept of “geodesic length” (though I’d be interested to know if this can be done rigorously).

      • physjam says:

        Actually I think I’ve been conceptually wrong in a very important respect – in both Euclidean geometry and physics there is no absolute notion of length. In both Euclidean space and physical space-time there is no meaning to a line segment of “length 1”; all that we can talk about is the ratio of lengths of two line segments, which is an affine notion.
        Since we have less symmetry than in an “affine metric space” we expect to have a larger invariance group; the extra transformations are dilations. The breaking of this symmetry corresponds to choosing units – which is another arbitrary convention – we choose a particular line segment (it used to be a platinum-iridium bar in Paris we called the meter), and a dilation is a change of units (e.g. to inches or light seconds).

        The only invariants in Euclidean geometry, a 2 dimensional space, are lines and circles (a line segment is that part of a line lying within some circle). A Euclidean transformation must map lines to lines and circles to circles. These transformations can be generated by: translations (2d),dilations (1d) and rotations(1d). There are also inversions, but this is equivalent to a rotation through 180 degrees. (Note that dilations must be the same in every direction to preserve circles).

        The only invariants in non-relativistic mechanics, a 4 dimensional space-time, are lines and spacial spheres. A transformation must map lines to lines and spheres to spheres. These transformations can be generated by: spacial translations (3d), spacial rotations (3d), spacial dilations (1d), time translations (1d), time dilations (1d), shears in the spatial directions i.e. Galilean boosts (3d) and … I suppose shears in the time direction t \to t + \vec{a} \cdot \vec{x} (3d). This last one doesn’t really fit in: I’ll need to think about what it is.

        In relativistic mechanics, a 4 dimensional space-time, there are lines and 4-dimensional hyperboloids. A transformation must map lines to lines and hyperboloids to hyperboloids. These transformations are generated by: space-time translations (4d), dilations (1d), symmetries of the hyperboloid (rotations (3d) and Lorentz boosts (3d)) (6d).

        Actually it could be interesting to allow mappings between lines and circles (Mobius transformations) – this would make the geometry projective…

  3. physjam says:

    So I’m pretty sure the shears in the time dimension aren’t in ordinary physics. What’s the extra structure? Lines of constant time must be mapped to lines of constant time: it’s a sort of foliation of space-time. It is interesting to look at Galilean boosts as a limit of Lorentz boosts – the appearance of this foliation seems serendipitous and in particular the 1st order correction is a complicated transformation.

    Also physics isn’t invariant under scale transforms, but there is no canonical choice of units.

  4. Michael says:

    Hey, I’ve been thinking about something. When we teach Maths IB, we talk about linear transformations, and their associated standard matrix. If we take these linear transformations to be from R^n to R^m (i.e. there is an ordered, orthonormal basis on both spaces), then the correspondence between a linear transformation and its standard matrix is immediate. However, more generally, if we are talking about linear transformations from V (an n-dimensional real vector space) to W (an m-dimensional real vector space), we still need to choose bases for both spaces in order to get a standard matrix (with respect to those bases). This seems to be the case more often when doing Differential Geometry. For example, the derivative of a smooth map between smooth manifolds at a point is just a linear map between tangent spaces. It isn’t until you choose coordinates for (a neighbourhood of) your point, as well as coordinates for (a neighbourhood of) the image of that point, that you get bases for your tangent spaces, and hence a standard matrix.

  5. Pingback: Geometry of division rings | physjam

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