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Lie Bracket and Torsion Tensor

In order to work with the Lie bracket, we need to consider the meaning of "travelling along" a vector field. Consider a vector field v=ex+xey\boldsymbol{v} = \boldsymbol{e_x} + x \boldsymbol{e_y}. This vector field tells us the velocity of a curve parametrized by λ\lambda:

ddλ=dxdλey+dydλey=ex+xey, \begin{align*} \frac{d}{d\lambda} &= \frac{d x}{d\lambda} \boldsymbol{e_y} + \frac{d y}{d\lambda} \boldsymbol{e_y} = \boldsymbol{e_x} + x \boldsymbol{e_y}, \end{align*}

this gives us the following partial differential equations with solutions:

dxdλ=1,dydλ=x, \begin{align*} \frac{d x}{d\lambda} &= 1, \\ \frac{d y}{d\lambda} &= x, \end{align*}

with the solutions:

x(λ)=λ+x0,y(λ)=12λ2+x0λ+y0, \begin{align*} x(\lambda) &= \lambda + x_0, \\ y(\lambda) &= \frac{1}{2} \lambda^2 + x_0 \lambda + y_0, \end{align*}

where x0x_0 and y0y_0 describe the initial position. These are equations for a flow curve. Flow curve is a curve tangent to all vectors in a vector field.

The Lie bracket, also called the commutator, is defined as follows:

[u,v]=u(v)−v(u),[\boldsymbol{u}, \boldsymbol{v}] = \boldsymbol{u}(\boldsymbol{v}) - \boldsymbol{v}(\boldsymbol{u}),

this tells us the difference between applying u\boldsymbol{u} to v\boldsymbol{v} and applying v\boldsymbol{v} to u\boldsymbol{u}. The Lie bracket may be expanded:

[u,v]=uμ∂∂xμ(vν∂∂xν)−vμ∂∂xμ(uν∂∂xν)=uμ(∂vν∂xμ∂∂xν+vν∂∂xμ∂∂xν)−vμ(∂uν∂xμ∂∂xν+uν∂∂xμ∂∂xν)=uμ∂vν∂xμ∂∂xν+uμvν∂∂xμ∂∂xν−vμ∂uν∂xμ∂∂xν−vμuν∂∂xμ∂∂xν=uμ∂vν∂xμ∂∂xν−vμ∂uν∂xμ∂∂xν=(uμ∂vν∂xμ−vμ∂uν∂xμ)eν. \begin{align*} [\boldsymbol{u}, \boldsymbol{v}] &= u^{\mu} \frac{\partial}{\partial x^{\mu}} \left(v^{\nu} \frac{\partial}{\partial x^{\nu}}\right) - v^{\mu} \frac{\partial}{\partial x^{\mu}} \left(u^{\nu} \frac{\partial}{\partial x^{\nu}}\right) \\ &= u^{\mu} \left(\frac{\partial v^{\nu}}{\partial x^{\mu}} \frac{\partial}{\partial x^{\nu}} + v^{\nu} \frac{\partial}{\partial x^{\mu}} \frac{\partial}{\partial x^{\nu}}\right) - v^{\mu} \left(\frac{\partial u^{\nu}}{\partial x^{\mu}} \frac{\partial}{\partial x^{\nu}} + u^{\nu} \frac{\partial}{\partial x^{\mu}} \frac{\partial}{\partial x^{\nu}}\right) \\ &= u^{\mu} \frac{\partial v^{\nu}}{\partial x^{\mu}} \frac{\partial}{\partial x^{\nu}} + u^{\mu} v^{\nu} \frac{\partial}{\partial x^{\mu}} \frac{\partial}{\partial x^{\nu}} - v^{\mu} \frac{\partial u^{\nu}}{\partial x^{\mu}} \frac{\partial}{\partial x^{\nu}} - v^{\mu} u^{\nu} \frac{\partial}{\partial x^{\mu}} \frac{\partial}{\partial x^{\nu}} \\ &= u^{\mu} \frac{\partial v^{\nu}}{\partial x^{\mu}} \frac{\partial}{\partial x^{\nu}} - v^{\mu} \frac{\partial u^{\nu}}{\partial x^{\mu}} \frac{\partial}{\partial x^{\nu}} \\ &= \left(u^{\mu} \frac{\partial v^{\nu}}{\partial x^{\mu}} - v^{\mu} \frac{\partial u^{\nu}}{\partial x^{\mu}}\right) \boldsymbol{e_{\nu}}. \end{align*}

As an example, consider the following vector fields:

u=ex,v=xey. \begin{align*} \boldsymbol{u} &= \boldsymbol{e_x}, \\ \boldsymbol{v} &= x \boldsymbol{e_y}. \end{align*}
Illustration of uIllustration of v

u\boldsymbol{u}

v\boldsymbol{v}

The Lie bracket is equal to:

[u,v]=(uμ∂vx∂xμ−vμ∂ux∂xμ)ex+(uμ∂vy∂xμ−vμ∂uy∂xμ)ey=(ux∂vx∂x−vy∂ux∂y)ex+(ux∂vy∂x−vy∂uy∂y)ey=ey. \begin{align*} [\boldsymbol{u}, \boldsymbol{v}] &= \left(u^{\mu} \frac{\partial v^x}{\partial x^{\mu}} - v^{\mu} \frac{\partial u^x}{\partial x^{\mu}}\right) \boldsymbol{e_x} + \left(u^{\mu} \frac{\partial v^y}{\partial x^{\mu}} - v^{\mu} \frac{\partial u^y}{\partial x^{\mu}}\right) \boldsymbol{e_y} \\ &= \left(u^x \frac{\partial v^x}{\partial x} - v^y \frac{\partial u^x}{\partial y}\right) \boldsymbol{e_x} + \left(u^x \frac{\partial v^y}{\partial x} - v^y \frac{\partial u^y}{\partial y}\right) \boldsymbol{e_y} \\ &= \boldsymbol{e_y}. \end{align*}

Looking at both vector fields in one plot:

Illustration of u and v

Blue: u\boldsymbol{u}

Red: v\boldsymbol{v}

(Not to scale)

We can make sense of this by considering the flow curves. We start at a point and travel along the blue vector field (u\boldsymbol{u}) and this changes the vector field v\boldsymbol{v}. Then we go back to the same point and travel along the red vector field (v\boldsymbol{v}) and this changes the vector field u\boldsymbol{u} (not in this specific case). The difference is the Lie bracket:

Illustration of u and v with the flow lines

Green: u(v)\boldsymbol{u}(\boldsymbol{v})

Orange: v(u)\boldsymbol{v}(\boldsymbol{u})

Black: [u,v]=u(v)−v(u)[\boldsymbol{u}, \boldsymbol{v}] = \boldsymbol{u}(\boldsymbol{v}) - \boldsymbol{v}(\boldsymbol{u})

(Not to scale)

When the lie bracket is zero, it means that the flow curves form a closed loop.

We may also say that the coordinate lines are the flow curves along basis vectors. And their Lie bracket is zero:

[eμ,eν]=∂∂xμ(∂∂xν)−∂∂xν(∂∂xμ)=∂2∂xμ∂xν−∂2∂xν∂xμ=0, \begin{align*} [\boldsymbol{e_{\mu}}, \boldsymbol{e_{\nu}}] &= \frac{\partial}{\partial x^{\mu}} \left(\frac{\partial}{\partial x^{\nu}}\right) - \frac{\partial}{\partial x^{\nu}} \left(\frac{\partial}{\partial x^{\mu}}\right) \\ &= \frac{\partial^2}{\partial x^{\mu} \partial x^{\nu}} - \frac{\partial^2}{\partial x^{\nu} \partial x^{\mu}} \\ &= 0, \end{align*}

and this is good. It means that the coordinate lines close.

The torsion tensor T(u,v)T(\boldsymbol{u}, \boldsymbol{v}) acting on the vectors u\boldsymbol{u} and v\boldsymbol{v} tells us the separation between parallel transported vectors:

Illustration of torsion tensor

where vp\boldsymbol{v_p} is parallel transport of v\boldsymbol{v} and up\boldsymbol{u_p} is parallel transport of u\boldsymbol{u}. The covariant derivative measures how much a vectors changes relative to the parallel transform of the vector. We don't know the parallel transported vector. But we can still express the torsion tensor as follows:

T(u,v)=∇uv−∇vu−[u,v].T(\boldsymbol{u}, \boldsymbol{v}) = \nabla_{\boldsymbol{u}} \boldsymbol{v} - \nabla_{\boldsymbol{v}} \boldsymbol{u} - [\boldsymbol{u}, \boldsymbol{v}].

Torsion-free property is a property of a connection, meaning it does not depend on the vector fields we choose. It means that the torsion tensor is zero:

T(u,v)=∇uv−∇vu−[u,v]=0.T(\boldsymbol{u}, \boldsymbol{v}) = \nabla_{\boldsymbol{u}} \boldsymbol{v} - \nabla_{\boldsymbol{v}} \boldsymbol{u} - [\boldsymbol{u}, \boldsymbol{v}] = \boldsymbol{0}.

We can solve for the components:

T(u,v)=∇uv−∇vu−[u,v]=uμ(∂vσ∂xμ+vνΓσμν)eσ−vμ(∂uσ∂xμ+uνΓσμν)eσ−(uμ∂vσ∂xμ−vμ∂uσ∂xμ)eσ=(uμ∂vσ∂xμ+uμvνΓσμν−vμ∂uσ∂xμ−vμuνΓσμν−uμ∂vσ∂xμ+vμ∂uσ∂xμ)eσ=(uμvνΓσμν−vμuνΓσμν)eσ=(uμvνΓσμν−uμvνΓσνμ)eσ=uμvν(Γσμν−Γσνμ)eσ, \begin{align*} T(\boldsymbol{u}, \boldsymbol{v}) &= \nabla_{\boldsymbol{u}} \boldsymbol{v} - \nabla_{\boldsymbol{v}} \boldsymbol{u} - [\boldsymbol{u}, \boldsymbol{v}] \\ &= u^{\mu} \left(\frac{\partial v^{\sigma}}{\partial x^{\mu}} + v^{\nu} \Gamma^{\sigma}{}_{\mu \nu}\right) \boldsymbol{e_{\sigma}} - v^{\mu} \left(\frac{\partial u^{\sigma}}{\partial x^{\mu}} + u^{\nu} \Gamma^{\sigma}{}_{\mu \nu}\right) \boldsymbol{e_{\sigma}} - \left(u^{\mu} \frac{\partial v^{\sigma}}{\partial x^{\mu}} - v^{\mu} \frac{\partial u^{\sigma}}{\partial x^{\mu}}\right) \boldsymbol{e_{\sigma}} \\ &= \left(u^{\mu} \frac{\partial v^{\sigma}}{\partial x^{\mu}} + u^{\mu} v^{\nu} \Gamma^{\sigma}{}_{\mu \nu} - v^{\mu} \frac{\partial u^{\sigma}}{\partial x^{\mu}} - v^{\mu} u^{\nu} \Gamma^{\sigma}{}_{\mu \nu} - u^{\mu} \frac{\partial v^{\sigma}}{\partial x^{\mu}} + v^{\mu} \frac{\partial u^{\sigma}}{\partial x^{\mu}}\right) \boldsymbol{e_{\sigma}} \\ &= \left(u^{\mu} v^{\nu} \Gamma^{\sigma}{}_{\mu \nu} - v^{\mu} u^{\nu} \Gamma^{\sigma}{}_{\mu \nu}\right) \boldsymbol{e_{\sigma}} \\ &= \left(u^{\mu} v^{\nu} \Gamma^{\sigma}{}_{\mu \nu} - u^{\mu} v^{\nu} \Gamma^{\sigma}{}_{\nu \mu}\right) \boldsymbol{e_{\sigma}} \\ &= u^{\mu} v^{\nu} (\Gamma^{\sigma}{}_{\mu \nu} - \Gamma^{\sigma}{}_{\nu \mu}) \boldsymbol{e_{\sigma}}, \end{align*}

so the components of the torsion tensor are:

Tσμν=Γσμν−Γσνμ,T^{\sigma}{}_{\mu \nu} = \Gamma^{\sigma}{}_{\mu \nu} - \Gamma^{\sigma}{}_{\nu \mu},

which only depend on the connection.

And again, the torsion-free property means that the torsion tensor components are zero:

Tσμν=Γσμν−Γσνμ=0,Γσμν=Γσνμ. \begin{align*} T^{\sigma}{}_{\mu \nu} = \Gamma^{\sigma}{}_{\mu \nu} - \Gamma^{\sigma}{}_{\nu \mu} &= 0, \\ \Gamma^{\sigma}{}_{\mu \nu} &= \Gamma^{\sigma}{}_{\nu \mu}. \end{align*}

We could also express the torsion tensor as follows:

Tσμν=2Γσ[μν],T^{\sigma}{}_{\mu \nu} = 2 \Gamma^{\sigma}{}_{[\mu \nu]},

where the brackets mean that the components are symmetrized. For any general tensor:

A[μ1μ2...μn]νσ=1n!(Aμ1μ2...μnνσ+alternating sum over permutations of μ indices),A_{[\mu_1 \mu_2 ... \mu_n] \nu}{}^{\sigma} = \frac{1}{n!} (A_{\mu_1 \mu_2 ... \mu_n \nu}{}^{\sigma} + \textrm{alternating sum over permutations of \(\mu\) indices}),

where alternating sum means minus for odd permutations and plus for even permutations.

For a general tensor, components may be symmetrized:

A(μ1μ2...μn)νσ=1n!(Aμ1μ2...μnνσ+sum over permutations of μ indices).A_{(\mu_1 \mu_2 ... \mu_n) \nu}{}^{\sigma} = \frac{1}{n!} (A_{\mu_1 \mu_2 ... \mu_n \nu}{}^{\sigma} + \textrm{sum over permutations of \(\mu\) indices}).

A tensor having symmetric or antisymmetric components means:

Aμν=Aνμsymmetric,Bμν=−Bνμantisymmetric. \begin{align*} A_{\mu \nu} &= A_{\nu \mu} & \textrm{symmetric}, \\ B_{\mu \nu} &= -B_{\nu \mu} & \textrm{antisymmetric}. \end{align*}

A tensor with symmetrized components A(μ1μ2...μn)νσA_{(\mu_1 \mu_2 ... \mu_n) \nu}{}^{\sigma} or antisymmetrized components A[μ1μ2...μn]νσA_{[\mu_1 \mu_2 ... \mu_n] \nu}{}^{\sigma} is not the same tensor as Aμ1μ2...μnνσA_{\mu_1 \mu_2 ... \mu_n \nu}{}^{\sigma}.

For a torsion free connection, the Christoffel symbols are the same as symmetrized:

Γσμν=Γσ(μν).\Gamma^{\sigma}{}_{\mu \nu} = \Gamma^{\sigma}{}_{(\mu \nu)}.