The Road to Reality - exercise on scalar product

[cianfa72]

This is a non-abelian Lie group. The tangent vectors (at the identity) are
$$X=\begin{pmatrix} 1&0\\0&-1 \end{pmatrix}\, , \,Y=\begin{pmatrix} 0&1\\0&0 \end{pmatrix}$$ with $[X,Y]=2Y.$

Ah ok, one gets the basis of tangent vectors at the Lie group identity (i.e. Lie algebra basis vectors) calculating the derivatives of group element w.r.t. the parameters $t$ and $c$ and evaluating them at $t=0,c=0$.

 

[cianfa72]

Sorry, regarding my previous statement

I suspect that the meaning of the expression $\nabla_{\nu} \nabla_{\mu}f$ is not actually $\nabla_{\nu}( \nabla_{\mu}f)$ but $(\nabla \nabla f)_{\nu \mu}$. Note that $\nu,\mu$ are indices (1,2,3...) in any local chart.

What is the actual meaning of $\nabla_{\nu} \nabla_{\mu}f$ ?
Thanks.

 

[martinbn]

Sorry, regarding my previous statement

What is the actual meaning of $\nabla_{\nu} \nabla_{\mu}f$ ?
Thanks.

The first derivative produces a one form, the second derivative is of that one form.

 

[cianfa72]

The first derivative produces a one form, the second derivative is of that one form.

I think it is matter of notation: $\nabla_{\nu}\nabla_{\mu}f$ should be written as $$(\mathbf {\nabla} (\mathbf {\nabla} f))_{\nu \mu}$$ i.e. starting from a scalar field $f$ take the covariant derivative to get its gradient (1-form) then apply the covariant derivative again. You get a (0,2) tensor. Fix a basis $\{\mathbf e_{\alpha}\}$ in the tangent space at any point and insert basis vectors in order into the (0,2) tensor "slots" (i.e. do the contractions). This way you get the $\nu,\mu$ components of the (0,2) tensor in that basis.

Now it makes sense what is going on in post #4: since on manifolds the Connection coefficients in general do not vanish, the covariant derivative of the gradient doesn't result in second-order mixed partial derivatives of $f$, hence the aforementioned expression doesn't commute in $\nu, \mu$.

 

[cianfa72]

Btw, even in $\mathbb R^n$ if one picks a non-standard affine connection (i.e. a non Levi-Civita connection), then $\nabla_{\nu} \nabla_{\mu}f$ doesn't commute as well.

 

[martinbn]

Btw, even in $\mathbb R^n$ if one picks a non-standard affine connection (i.e. a non Levi-Civita connection), then $\nabla_{\nu} \nabla_{\mu}f$ doesn't commute as well.

As long as the torsion is not zero.

 

[cianfa72]

Just for practice I did the explicit calculation. From here covariant derivative start from a 1-form $\theta = \theta_{\eta}dx^{\eta}$ written in the chart with coordinates $\{ x^{\eta} \}$.

The gradient $\nabla f$ is by definition the 1-form/covector field $\nabla f = \left ( \partial_{\mu} f \right ) dx^{\mu}$. Hence $$\nabla_{\nu} \nabla_{\mu} f = (\nabla \nabla f)_{\nu \mu} = \left ( \nabla ((\partial_{\eta}f) dx^{\eta}) \right )_{\nu \mu}= \left ( \partial_{\nu} \partial_{\mu}f - (\partial_{\eta}f)\Gamma^{\eta}_{ \mu \nu} \right ) dx^{\nu} \otimes dx^{\mu}$$
From the above we can see that whether the connection $\Gamma$ is symmetric in the lower indices (or vanish) in such coordinates (hence in all coordinates) then $\nabla_{\nu} \nabla_{\mu} f = \nabla_{\mu} \nabla_{\nu} f$ i.e. they commute.