Tensors Defined on Differentiable Manifolds 📂Geometry

Tensors Defined on Differentiable Manifolds

Definition¹

Let $M$ be a $n$-dimensional differentiable manifold, $\mathcal{D}(M)$ be the set of differentiable functions on $M$, $\mathfrak{X}(M)$ be the set of all vector fields on $M$.

$$ \mathcal{D}(M) := \left\{ \text{all real-valued functions of class } C^{\infty} \text{ defined on } M \right\} $$

$$ \frak{X}(M) := \left\{ \text{all vector fileds of calss } C^{\infty} \text{ on } M \right\} $$

A multilinear function $T$ as follows is called a $r$-order tensor.

$$ T : \overbrace{\frak{X}(M) \times \cdots \times \frak{X}(M)}^{r} \to \mathcal{D}(M) $$

Explanation

$\frak{X}(M)$ becomes a module over $\mathcal{D}(M)$. By definition, the following holds. For all $X, Y \in \frak{X}(M),\ f,g\in \mathcal{D}(M)$,

$$ T(Y_{1}, \dots, fX + gY, \dots, Y_{r}) = fT(Y_{1}, \dots, X, \dots, Y_{r}) + gT(Y_{1}, \dots, Y, \dots, Y_{r}) $$

The characteristic of tensors is that they depend not on the coordinate system, but only on the value at each point. To illustrate this, fix the point $p \in M$ and consider the vector field $E_{i}, \dots, E_{n} \in \frak{X}(M)$ that makes $\left\{ E_{i}(p) \right\}$, which is near $p$, the basis of the tangent space $T_{p}M$. This $\left\{ E_{i} \right\}$ is called a moving frame on $U$. Now, let’s represent the reduction mapping of vector fields $Y_{i}$ onto $U$ with the moving frame $\left\{ E_{i} \right\}$ as follows.

$$ Y_{1} = \sum_{i_{1}}y_{i_{1}}E_{i_{1}},\quad \dots,\quad Y_{r} = \sum_{i_{r}}y_{i_{r}}E_{i_{r}} $$

And let’s think of other vector fields $\left\{ Z_{j} = \sum z_{k_{j}}E_{k_{j}} \right\} \subset \frak{X}(M)$ that are ‘only’ equal in value at the point $p$ to $Y_{i}$.

$$ \begin{align*} && Z_{j}(p) &= Y_{j}(p) \\ \implies && z_{k_{j}}(p)E_{k_{j}}(p) &= y_{k_{j}}(p)E_{k_{j}}(p) \\ \implies && z_{k_{j}}(p) &= y_{k_{j}} \end{align*} $$

Then the following is obtained.

$$ \begin{align*} T(Y_{1}, Y_{2}, \dots, Y_{n})(p) &= y_{i_{1}}(p)\cdots y_{i_{r}}(p) T(E_{i_{1}}(p), \dots, E_{i_{r}}(p)) \\ &= z_{i_{1}}(p)\cdots z_{i_{r}}(p) T(E_{i_{1}}(p), \dots, E_{i_{r}}(p)) \\ &= T(Z_{1}, Z_{2}, \dots, Z_{n})(p) \end{align*} $$

Therefore, $T(Y_{1}, Y_{2}, \dots, Y_{n})(p)$ depends only on the value of $Y_{i}$ at $p$, and not on the coordinate system.

Examples

Curvature Tensor

The Riemann curvature $R$, defined as follows, is a $4$-order tensor.

$$ R : \frak{X}(M) \times \frak{X}(M) \times \frak{X}(M) \times \frak{X}(M) \to \mathcal{D}(M) $$

$$ R(X, Y, Z, W) = \left\langle R(X, Y)Z, W \right\rangle, \quad X, Y, Z, W \in \frak{X}(M) $$

Regarding the moving frame $\left\{ X_{i} = \dfrac{\partial }{\partial x_{i}} \right\}$,

$$ R(X_{i}, X_{j}, X_{k}, X_{l}) = R_{ijkl} $$

Metric Tensor

$$ g : \frak{X}(M) \times \frak{X}(M) \to \mathcal{D}(M) $$

$$ g(X, Y) = \left\langle X, Y \right\rangle, \quad X, Y \in \frak{X}(M) $$

The Riemannian metric $g$ is a $2$-order tensor.

Connection

$$ \nabla : \frak{X}(M) \times \frak{X}(M) \times \frak{X}(M) \to \mathcal{D}(M) $$

$$ \nabla(X, Y, Z) = \left\langle \nabla_{X}Y, Z \right\rangle, \quad X, Y, Z \in \frak{X}(M) $$

The Levi-Civita connection $\nabla$, defined as above, is not a tensor since it is not linear with respect to the $Y$ components.