Torsion and Curvature

Torsion

Torsion

Let $M$ be a manifold with an affine connection $\nabla$ on the tangent bundle. The torsion tensor of $\nabla$ is the vector valued $2$-form defined on vector fields $X$ and $Y$ by $T(X,Y)={\nabla }_{X}Y-{\nabla }_{Y}X-[X,Y]$where $[X,Y]$ is the Lie bracket of $X$ and $Y$.We say an affine connection is torsion free if its torsion tensor vanishes identically.

Proposition

The torsion $T$ is tensorial.

Proof Let $X,Y$ be vector fields and $f$ a smooth function. Then we have $$\begin{aligned}T(fX,Y)&=\nabla_{fX}Y-\nabla_Y(fX)-[fX,Y]\&=f\nabla_XY- Y(f)X-f\nabla_YX+Y(f)X-f[X,Y]\&=f(\nabla_XY-\nabla_YX-[X,Y])\&=fT(X,Y)\end{aligned}$$$\square$

Proposition

In a local coordinate system, the torsion tensor is given by $T_{ij}^k={\Gamma }_{ij}^k-{\Gamma }_{ji}^k$

Proof $\begin{array}{r}T({\partial }_i,{\partial }_j)={\nabla }_i{\partial }_j-{\nabla }_j{\partial }_i-[{\partial }_i,{\partial }_j]=({\Gamma }_{ij}^k-{\Gamma }_{ji}^k){\partial }_k\end{array}$It follows that $T_{ij}^k={\Gamma }_{ij}^k-{\Gamma }_{ji}^k$. $\square$

Corollary

An affine connection is torsion free if and only if the first two components of the Christoffel symbols are symmetric. i.e. ${\Gamma }_{ij}^k={\Gamma }_{ji}^k$.

Proof It is straightforward in coordinate form. $\square$

Intension for Curvature

1. Connection and the Local Gauge Potential

A connection ${\nabla }_X:\Gamma (E)\to \Gamma (E)$ prescribes how to take directional derivatives of sections along a vector field $X\in \mathfrak{X}(M)$. Over a local trivialization $E{|}_U\cong U\times {\mathbb{R}}^k$, the connection splits into a flat (trivial) differentiator $d$ and a local correction term $A$: $\nabla =d+A$where $A\in {\Omega }^1(U,\mathfrak{gl}(k,\mathbb{R}))$ is the connection 1-form (Gauge Potential). Under a change of frame $g_{\alpha \beta }:U_{\alpha }\cap U_{\beta }\to \text{GL}(k,\mathbb{R})$, $A$ transforms affinely via an extra gauge term: $A_{\alpha }=g_{\alpha \beta }{A}_{\beta }{g}_{\alpha \beta }^{-1}-(d{g}_{\alpha \beta })g_{\alpha \beta }^{-1}$ Consequently, $A$ is not a global tensor; its local profile is highly dependent on the choice of gauge.

2. Curvature as the Intrinsic Tensor

Curvature $F_{\nabla }$ measures the failure of the connection to commute, acting as an obstruction to local flatness: $F_{\nabla }(X,Y)={\nabla }_X{\nabla }_Y-{\nabla }_Y{\nabla }_X-{\nabla }_{[X,Y]}$Computed from the local gauge potential via the structure equation: $F=dA+A\wedge A\in {\Omega }^2(U,\mathfrak{gl}(k,\mathbb{R}))$ The non-tensorial derivative terms of $A$ cancel perfectly under a frame change, leaving a clean, linear transformation:$F_{\alpha }=g_{\alpha \beta }{F}_{\beta }{g}_{\alpha \beta }^{-1}$ Thus, $F_{\nabla }\in {\Omega }^2(M,\text{End}(E))$ is a true global tensor (Gauge Field Strength). It captures the physical, coordinate-independent reality of the field that cannot be “gauged away.”

3. Global Affine Space and Physical Dynamics

The space of all connections ${\mathcal{A}}_E$ is not a vector space, but an affine space modeled on the vector space of global endomorphism-valued 1-forms: ${\mathcal{A}}_E={\nabla }_0+{\Omega }^1(M,\text{End}(E))$Yang-Mills Functional: In physics, the curvature tensor acts as the field strength, and its total kinetic energy yields the Yang-Mills action functional: $\mathcal{Y}\mathcal{M}(A)=\Vert F_A{\Vert }_{L^2}^2$When $E=TM$ and $\nabla ={\nabla }^{\text{L.C.}}$ (the canonical Levi-Civita connection determined by the metric $g$), this curvature specializes to the classical Riemann Curvature Tensor: $R:=F_{{\nabla }^{\text{L.C.}}}\in {\Omega }^2(M,\text{End}(TM))\cong \Gamma (M,{\Lambda }^2{T}^{\ast }M\otimes T^{\ast }M\otimes TM)$

Curvature

Riemann Curvature

Let $M$ be a manifold with an affine connection $\nabla$ on the tangent bundle. The (Riemann) curvature tensor of $\nabla$ is the vector valued $2$-form defined on vector fields $X$ and $Y$ by $R(X,Y):={\nabla }_X{\nabla }_Y-{\nabla }_Y{\nabla }_X-{\nabla }_{[X,Y]}.$ Alternatively, the Riemann tensor can be defined as a $(3,1)$-tensor that acts on $X,Y,Z\in \mathfrak{X}(M)$ and $\omega \in {\Lambda }^1(M)$ by $R(X,Y,Z,\omega ):=\omega \left({\nabla }_X{\nabla }_{Y}Z-{\nabla }_{}Y{\nabla }_{X}Z-{\nabla }_{[X,Y]}Z\right).$

In fact, we can also introduce the curvature tensor to any vector bundle:

Curvature Tensor

Suppose $\nabla$ is a connection on a vector bundle $E$ over $M$, then the curvature tensor $R$ of $\nabla$ is the section of $T^{\ast }M\otimes T^{\ast }M\otimes E^{\ast }\otimes E$ defined by $R(X,Y,\xi ):={\nabla }_Y({\nabla }_X\xi )-{\nabla }_X({\nabla }_Y\xi )-{\nabla }_{[Y,X]}\xi .$

Geometric Meaning of Curvature

The Riemann curvature tensor gives a measure of the non-commutativity of covariant derivatives acting on tensor fields on $M$.

Proposition

An affine connection is flat if and only if there exist a local coordinate system in which Christoffel symbols vanish.

Proof

Theorem

An affine connection is flat if and only if the parallel transport is path independent.

Proposition

The components of the Riemann curvature tensor (as a (3,1)-tensor) are given by $R({\partial }_{x_i},{\partial }_{x_j}){\partial }_{x_k}={\sum }_l{R}_{ijk}^l{\partial }_{x_l}$ $R_{ijk}^l={\partial }_{x_i}{\Gamma }_{jk}^l-{\partial }_{x_j}{\Gamma }_{ik}^l+{\sum }_m\left({\Gamma }_{im}^l{\Gamma }_{jk}^m-{\Gamma }_{jm}^l{\Gamma }_{ik}^m\right)$

Musical Isomorphism

Proposition

Equivalently, we can express Riemannian curvature as a $(4,0)$ tensor: $R_{ijkl}={\sum }_m{R}_{ijk}^m{g}_{ml},$ and we have $R_{ijkl}=-R_{jikl}=-R_{ijlk}$, $R_{ijkl}=R_{klij}$.

Curvature as an $\mathfrak{so}(n)$-Valued 2-Form

Let $(M,g)$ be an $n$-dimensional Riemannian manifold equipped with the Levi-Civita connection $\nabla$. Let $U\subset M$ be a local coordinate neighborhood with a local orthonormal frame. Then, the connection 1-form $A\in {\Omega }^1(U,\mathfrak{gl}(n,\mathbb{R}))$ satisfies the metric compatibility condition: $A=-A^T$. Consequently, the local connection 1-form takes values in the orthogonal Lie algebra, $A\in {\Omega }^1(U,\mathfrak{so}(n))$. Since the Riemann curvature tensor $R$ can be expressed locally via the second structural equation as: $R=dA+A\wedge A$ it follows that $R$ is a $\mathfrak{so}(n)$-valued 2-form on $M$. Globally, this identifies the curvature tensor as a section: $R\in {\Omega }^2(M,\mathfrak{so}(M))\subset {\Omega }^2(M,\text{End}(TM))$

Bianchi identity

1st Bianchi identity: $R(X,Y)Z+R(Y,Z)X+R(Z,X)Y=0$ In local coordinates, this translates to the cyclic permutation of the first three indices: $R_{ijk}^l+R_{jki}^l+R_{kij}^l=0$ 2nd Bianchi identity: ${\nabla }_{X}R(Y,Z)+{\nabla }_{Y}R(Z,X)+{\nabla }_{Z}R(X,Y)=0$ In local coordinates, this is expressed using the semicolon notation for covariant differentiation: $R_{ijk;m}^l+R_{ijm;k}^l+R_{ijk;m}^l=0⟹R_{ijk;m}^l+R_{ijm;k}^l+R_{imk;j}^l=0$

Sectional Curvature

Let $p\in M$ and let $\sigma \subset T_{p}M$ be a 2-dimensional subspace (a plane section) of the tangent space, so that $\sigma \in {\text{Gr}}_2(T_{p}M)$. If $\sigma$ is spanned by two linearly independent vectors $X,Y\in T_{p}M$, the sectional curvature $K(\sigma )$ of $\sigma$ is defined by: $K(\sigma )=\frac{R(X,Y,X,Y)}{\Vert X{\Vert }^2\Vert Y{\Vert }^2-g(X,Y{)}^2}$ where $R(X,Y,X,Y)=g(R(X,Y)X,Y)$. The value $K(\sigma )$ is independent of the choice of the basis $\{X,Y\}$ for the plane $\sigma$. If $\{{\partial }_{x_i},{\partial }_{x_j}\}$ is a local coordinate basis, the components are given by: $K({\partial }_{x_i}\wedge {\partial }_{x_j})=\frac{R_{ijij}}{g_{ii}{g}_{jj}-(g_{ij}{)}^2}$

Ricci Curvature

Ricci tensor is the contraction of the Riemann curvature tensor over the the second and fourth indices (equivalently, first and fourth indices). Explicitly, it is a $(2,0)$-tensor such that$Ri{c}_{ij}={\sum }_k{R}_{ikjk}(p)$. For any $X,Y\in T_{p}M$, it is defined coordinate-free as the trace of the linear endomorphism $Z\mapsto R(Z,X)Y$: $\text{Ric}(X,Y)=\text{Tr}(Z\mapsto R(Z,X)Y)$ In local coordinates, the components ${\text{Ric}}_{ij}=\text{Ric}({\partial }_{x_i},{\partial }_{x_j})$ are given by contracting with the inverse metric tensor $g^{kl}$: ${\text{Ric}}_{ij}=R_{ikjl}{g}^{kl}={\sum }_k{R}_{ikj}^k$

Proposition

The Ricci tensor is symmetric.

Proof $R_{ij}=R_{ikj}^k=$

We can further contract the Ricci tensor to obtain the Ricci scalar. But since the Ricci Tensor is already purely covariant, one cannot contract it further without introducing a metric. So we have the following definition:

Ricci Scaler

Ricci scaler is a $(0,0)$-tensor, i.e. a smooth function, defined on a semi Riemannian manifold by $S_g={\sum }_{i,j}{g}^{ij}Ri{c}_{ij}$