Often in engineering physics, different vector spaces are used to visualize the trajectories (evolution) of systems. An example being the $6n$ dimensional phase space of $n$ particles. It is not very clear to me if this space is Euclidean or has a meaning of distance. I understand the 2-norm when the physical quantities are the same (e.g., only positions or only momenta) but not when a vector has both. However even this seems to be a problem with some other spaces. For example if I consider the vector space of chemical composition of chemical system (mole numbers of all chemical species). I don't understand what "length" of a vector would signify?

  • $\begingroup$ "Vector space of chemical composition"? Why would it be a vector space? does it have addition subtraction multiplication cross product dot product etc that define a vector space? $\endgroup$
    – anna v
    Commented Aug 17, 2011 at 13:48
  • $\begingroup$ I think it has the properties of addition, subtraction, associativity, distributivity, etc. Although not an additive inverse. It is constrained to R+. I guess that makes it not a vector space. You are right. Thanks. $\endgroup$
    – Sankaran
    Commented Aug 17, 2011 at 17:10

3 Answers 3


The spaces you have in mind are in general not vector spaces. But it still makes sense to ask whether there is a useful concept of length around.

In particular, phase space (the 6n-dimensional space of positions and momenta) is a vector space if the underlying position space of single particle itself is. More generally, the particles would be constrained to move only on some subspace (e.g. a sphere) and the phase space would be cotantgent bundle of the $n$-fold product of the single-particle subspace. More generally still, the total position space need not have a product structure (which can happen if position of some particle depends on position of others). Again, the full space would be a cotangent bundle to the total position space.

Now, the phase space doesn't support a canonical notion of distance. Nevertheless, it does contain something else that is special: a symplectic two-form $\omega$ (the reason it's there is that phase space is always formed as a cotangent bundle) which turns it into symplectic manifold. Using $\omega$, one can form a $2N$-form (N here is the dimension of the total position space) $\omega^n = \underbrace{\omega \wedge \cdots \wedge \omega}_{n \text{ times}}$. Since this is a form of a maximal degree we can use it to measure volumes and this leads to famous Liouville's theorem that states that Hamiltonian flows on the symplectic manifold are incompressible (i.e. evolution conserves phase volume).

This is all the structure we get for phase spaces. You can impose a metric on any manifold to turn it into a Riemannian manifold but this won't be canonical (i.e. there are many different ways to measure distances). In case your phase space happens to be a simple $6n$-dimensional vector space, you can use the Euclidean notion of a distance but, again, this won't be canonical.

In general, there is no canonical way to measure things and you simply shouldn't be expecting to be able to do so unless you are explictly given something nice like Euclidean space or a submanifold thereof.

  • $\begingroup$ Thanks for the detailed response. I was thinking completely independent (free) particles with no interactions (e.g., an ideal gas) but that also makes it a product n-fold product of a single-particle subspace. Might it have a Euclidean length ? (although uninteresting). However, I understand comments. I do not understand much about the two-form, but I will read about it. Is there an more accessible text for engineers that comes to mind? $\endgroup$
    – Sankaran
    Commented Aug 17, 2011 at 17:18
  • $\begingroup$ Also, if the single particle is constrained, is the manifold a subspace in the first place? In a general case that may not be true either would it? $\endgroup$
    – Sankaran
    Commented Aug 17, 2011 at 17:25
  • $\begingroup$ +1; fantastically informative answer! Finally a sense of what a symplectic manifold dawns onto me... $\endgroup$
    – Slaviks
    Commented Aug 17, 2011 at 18:54
  • $\begingroup$ @Sankaran: the classical text that I always recommend is Fecko, see my answer here - physics.stackexchange.com/questions/1601/… ; As for the single particle, it is a subspace for all reasonable constraint forces. In general you could of course come up with pathological subspaces which wouldn't be manifolds (e.g. fractals) but these don't occur naturally. $\endgroup$
    – Marek
    Commented Aug 18, 2011 at 6:38
  • $\begingroup$ Thanks Marek. I will look into this book, as also the one by V.I. Arnold. $\endgroup$
    – Sankaran
    Commented Aug 18, 2011 at 20:37

There are cases where a metric on a phase space is needed\useful. I'll describe to you two cases having very interesting physics.

Actually, the example given in the question is an example of the first case. Please notice that an Eucledian metric of the form $p^2 + q^2$ can be interpretted as the total (kinetic + potential) energy of a harmonic oscillator.

More rigorously, when the phase space is a Kahler manifold, there are compatible metric, closed symplectic form and a complex structure: (A Hermitian manifold is a complex manifold equipped with a tensor $ h = h_{\alpha \bar{\beta}}dz^{\alpha}d\bar{z}^{\beta}$, such that the matrix elements $ h_{\alpha \bar{\beta}}$ are the elements of a Hermitian matrix).The symplectic form $\omega = \mathrm{Im}(h)$ is the imaginary part of the Hermitian form, the metric $g = \mathrm{Re}(h)$ is the real part and the complex structure $J$, satisfies $\omega (X,Y) = g(JX,Y)$ ($X$, $Y$ are vector fields). ( A Kahler manifold is a Hermitian manifold with a closed two form $\omega$)

Let's consider the example of the Eucledian two dimensional phase space, Here, we can define $z = x+ip$, then:

$\omega = dz \wedge d\bar{z}$

$g = \frac{1}{2} (dz \otimes d\bar{z} + d\bar{z}) $

These types of spaces admit holomorphic or Kahler quantization, in our example, the quantization space is the Bargmann space of $L^2$ holomorphic functions:

$\langle \psi, \phi \rangle = \int \overline{\psi(z)} \phi(z) \mathrm{exp}(-\bar{z} z) dz d\bar{z}$

Please observe again that the argument of the exponential is minus the Eucledian distance on the phase space. This example is sometimes referred to as the coherent state quantization of the harmonic oscillator. Here the energy eigenstates are just the monomials in $z$. One can return to the standard representation by the means of the (inverse) Bargmann transform.

A second application where a metric on a phase space is needed is when ones needs to put fermions (spinors) on the phase space (In this case one needs a spin structure as well). One easy way to see that is throug the anti-commutation relation of the Gamma matrices on a general (curved) manifold

$\{\gamma^{\mu},\gamma^{\nu}\} = g^{\mu\nu}$.

Thus the Dirac operator requires a metric. Dirac operators on phase spaces are used because their zero modes provide a way to define a quantization Hilbert space. In the above example, if we merge the Gaussian factor of the integration into the definition f wave function:

$\psi ^{\prime} (z, \bar{z}) = \psi(z) exp(\frac{-\bar{z}z}{2})$

The Hilbert space would not contain the Gaussian weighting:

$\langle \psi, \phi \rangle = \int \overline{\psi^{\prime}(z, \bar{z})} \phi^{\prime}(z, \bar{z}) dz d\bar{z}$

but now the wavefunctions are not holomorphic but rather satisfy the Dirac equation (in a uniform magnetic field):

$(\bar{\partial} + \frac{z}{2})\psi = 0$.

This is an almost trivial example of the use of the Dirac operator in quantization. For a good reference on the role of the metric on phase spaces and its connection to Dirac operators and quantization, please see the following article: "Quantization of systems with a general phase space equipped with a Riemannian metric" by Alicki and Klauder (sorry that I could not find an open version).

It is worthwhile to mention that the requirement of spin structure doesn't constitute a big constraint in the case of Kahler manifolds, since they can be equipped with a spin-c structure with witch a Dirac operator can be defined.

  • $\begingroup$ Hm, good of you to bring up Kähler manifolds, I have to give +1 for the shear amount of information present in this answer. On the other hand, it seems to be quite distant from the original purposes of the OP. $\endgroup$
    – Marek
    Commented Aug 18, 2011 at 11:14
  • $\begingroup$ Thanks David, although I only a small part of your detailed explanation (due to the lack of theoretical physics background), I do get some idea. Also, the harmonic oscillator is better example than the single free-particle example that I was thinking. $\endgroup$
    – Sankaran
    Commented Aug 18, 2011 at 20:42

Distances are measured via a metric and this is a symmetric bilinear form. One could ask can a useful distance could be constructed out of an antisymmetric blinear form, what we might call a cometric. In fact, they actually have been useful and are called symplectic forms. They are not used to measure distances but can be used to construct a volume form just like for a metric and in many cases this is just as useful.

Now phase space is the cotangent space of the configuration space and as cotangent spaces are naturally symplectic spaces, phase space is naturally symplectic. Thus it naturally carries a notion of volume. This, as already has been noted, is useful in forming Liouville's theorem.

More, in the metric case we can construct the Riemann curvature tensor field and this measures curvature from point to point. However, because of Darboux's formula in the cometric case, there is no such analogue of the curvature tensor here. In other words, there is no useful analogue of curvature for a symplectic manifold.


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