# does a vector field and its dual 1-form over an n-sphere form the heisenberg group?

Apologies, I don't claim my reasoning is perfect, but I would appreciate any critiques. Thank you.

Let us consider commutation relations on a general Riemannian manifold M , where the commutator is defined as:

$$\left[A,B\right]=AB-BA$$

Specifically, we consider the basic commutation relation between one of the set of basis on $M$ , $e^{\mu}$ and the basis dual to it, $e_{\mu}$ .$$\left[e^{\mu},e_{\mu}\right]$$

Choosing our basis to locally be coordinate basis we obtain: $$\left[e^{\mu},e_{\mu}\right]\xi=\left[\frac{\partial}{\partial x^{\mu}},dx\right]\xi$$

where $$\xi$$ is some test function defined over the manifold . Expanding out the terms of the commutator, we obtain:

$$\left[\frac{\partial}{\partial x^{\mu}},dx^{\mu}\right]\xi=\frac{\partial}{\partial x^{\mu}}\left(dx^{\mu}\xi\right)-dx^{\mu}\frac{\partial\xi}{\partial x^{\mu}}$$

$$=\frac{\partial(dx^{\mu})}{\partial x^{\mu}}\xi+\left(dx^{\mu}\frac{\partial\xi}{\partial x^{\mu}}-dx^{\mu}\frac{\partial\xi}{\partial x^{\mu}}\right)$$

$$=\frac{\partial(dx^{\mu})}{\partial x^{\mu}}\xi$$

written without the test function:

$$\left[\frac{\partial}{\partial x^{\mu}},dx^{\mu}\right]=\frac{\partial(dx^{\mu})}{\partial x^{\mu}}$$

The lattermost term, $\frac{\partial(dx^{\mu})}{\partial x^{\mu}}$ is the change in the element $dx^{\mu}$ with the change in $x^{\mu}$ . Because $dx^{\mu}$ is, in general, a function of position, it can be nonzero. Clearly such a quantity is related to curvature.

Because $dx^{\mu}=e^{\mu}$ is the dual to $\frac{\partial}{\partial x^{\mu}}=e_{\mu}$ , a change in one effects a change in another. Consider now Euler's definition of extrinsic curvature:

$$\frac{\partial(T_{x^{\mu}}M)}{\partial S}=\frac{1}{R}$$

Where R is the radius of curvature of the manifold in the $x^{\mu}$ direction, $T_{x^{\mu}}M$ is the unit tangent vector on M in the $\mu$ direction and $\partial S$ is the differential distance element. Of course the cotangent/dual space represented by the dual basis is defined as $$dx^{\mu}=(T_{x^{\mu}}M)^{\star}$$

Where $\star$ is the Hodge star operator. It is then evident that, for our local coordinate patch:

$$(T_{x^{\mu}}M)^{\star}=-i(T_{x^{\mu}}M)$$

Thus we can write:

$$\frac{\partial(dx^{\mu})}{\partial x^{\mu}}=\frac{\partial(T_{x^{\mu}}M)^{\star}}{\partial S}\left(\frac{\partial S}{\partial x^{\mu}}\right)=-i\frac{\partial(T_{x^{\mu}}M)}{\partial x^{\mu}}\left(\frac{\partial S}{\partial x^{\mu}}\right)=-\frac{i}{R}\left(\frac{\partial S}{\partial x^{\mu}}\right)$$

Now the distance element can be defined as:

$$dS=\left(dx^{\mu}dx_{\mu}\right)^{\frac{1}{2}}=\left(dx^{\mu}g_{\mu\nu}dx^{\nu}\right)^{\frac{1}{2}}=\gamma_{\mu}dx^{\mu}=\gamma^{\mu}dx_{\mu}$$

Where in the latter two term we have utilized Dirac's “trick” Interestingly, Dirac's mathematical “trick” of factoring the square root with gamma matrices can be used throughout classical relativity though, in practice, it is reserved exclusively for use in quantum mechanics for reasons ambiguous to the author.and factored the square root with gamma matrices defined by the relation: $\{\gamma^{\mu},\gamma^{\nu}\}=\gamma^{\mu}\gamma^{\nu}+\gamma^{\nu}\gamma^{\mu}=2g^{\mu\nu}$ .

Thus we may write the expression:

$$\frac{dS}{dx^{\mu}}=\gamma_{\mu}$$

Such that we may rewrite (7) as:

$$\left[\frac{\partial}{\partial x^{\mu}},dx^{\mu}\right]=\frac{\partial(dx^{\mu})}{\partial x^{\mu}}=-\frac{i}{R}\gamma_{\mu}$$

It is important to note here that equation (8) Equation (8) is rather fundamental, it demonstrates one may write the curvature in a coordinate direction in terms of a tangent vector and its dual 1-form. Note also that, in general, R and $\gamma_{\mu}$ are functions of the manifold coordinates.

Let us now consider a Taylor series expansion of $dx^{\mu}$ about the point $x_{0}$ .

$$dx^{\mu}=\sum_{n=0}^{\infty}\left(\frac{\partial^{n}(dx^{\mu})}{(\partial x^{\mu})^{n}}\mid_{x_{0}}\right)(x-x_{0})^{n}$$

$$=dx^{\mu}\mid_{x_{0}}+\left\{ \frac{\partial(dx^{\mu})}{\partial x^{\mu}}\mid_{x_{0}}\right\} (x^{\mu}-x_{0}^{\mu})+\cdots$$

$$=dx^{\mu}\mid_{x_{0}}-\frac{i}{R}\gamma_{\mu}\mid_{x_{0}}(x^{\mu}-x_{0}^{\mu})+\cdots$$

We will only consider terms to first order. This is equivalent to claiming the curvature is constant, or rather changes in curvature are negligible for the portion of the manifold we are considering. Inserting (12) back into expression (9) we obtain:

$$\left[\frac{\partial}{\partial x^{\mu}},dx^{\mu}\mid_{x_{0}}-\frac{i}{R}\gamma_{\mu}(x^{\mu}-x_{0})\right]\backsimeq-\frac{i}{R}\gamma_{\mu}$$

All constant terms cancel, yielding:

$$\left[\frac{\partial}{\partial x^{\mu}},-\frac{i}{R}\gamma_{\mu}x^{\mu}\right]=\left[\frac{\partial}{\partial x^{\mu}},-\frac{i}{R}\gamma^{\mu}x_{\mu}\right]=\backsimeq-\frac{i}{R}\gamma_{\mu}$$

This may be rewrittion equivalently as:

$$\left[-\frac{i}{R}\gamma^{\mu}\frac{\partial}{\partial x^{\mu}},\,x_{\mu}\right]\backsimeq-\frac{i}{R}\gamma_{\mu}$$

Equation (17) merits careful consideration. It is the first order approximation to curvature at a point varied from the point $x_{0}$ , but it is more than that, If the manifold being considered is an n-sphere all approximations are exact and (15) appears to form the Heisenberg group over it. Is this right or horribly wrong?

• Another way to ask this would be: For a manifold of constant curvature, do the elements of the tangent and cotangent space form the heisenberg group? – R. Rankin Apr 14 '16 at 19:12