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Could anybody explain to me the following:

If $$T_{ij}=\nabla_i V_j-\nabla_j V_i$$ where $V_i$ is a covector field and $\nabla_i$ is the covariant derivative, then $T_{ij}$ is a tensor field.

I am new to all hese covector, covariant, etc. and am very confused.

I believe the definition of a tensor is one which transforms according to the tensor transformation rule... right? But I don't see how that applies here.

Thank you in advance.

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Why wouldn't the transformation law apply? If you prove $T_{ij}$ obeys that law, you prove it to be a tensor. –  Muphrid Jan 22 '13 at 15:08
    
@Muphrid : I don't understand how to apply that to this case -- with the covariant derivatives and stuff... Would you mind explaining please? –  Nathaniel Jan 22 '13 at 15:14

3 Answers 3

up vote 2 down vote accepted

Here's how to check it if your connection is torsion-free (e.g. you're using the Levi-Civita connection).

First, note that \begin{align} T_{ij} & = \nabla_i V_j - \nabla_j V_i \\ & = (\partial_i V_j - \Gamma^k_{\phantom{k}ji} V_k) - (\partial_j V_i - \Gamma^k_{\phantom{k}ij} V_k) \\ & = \partial_i V_j - \partial_j V_i - \Gamma^k_{\phantom{k}ij}V_k + \Gamma^k_{\phantom{k}ij} V_k \\ & = \partial_i V_j - \partial_j V_i, \end{align} where the $\Gamma^k_{\phantom{k}ij}$ are the Christoffel symbols of the connection and $\Gamma^k_{\phantom{k}ij} = \Gamma^k_{\phantom{k}ji}$ by the assumption that the connection is torsion-free.

Now suppose we make a change of coordinates $x_i \mapsto x_{i'}$. Then \begin{align} T_{i'j'} & = \partial_{i'} V_{j'} - \partial_{j'} V_{i'} \\ & = \partial_{i'} \left( \frac{\partial x_k}{\partial x_{j'}} V_k \right) - \partial_{j'} \left( \frac{\partial x_l}{\partial x_{i'}} V_l \right) \\ & = \frac{\partial^2 x_k}{\partial x_{i'} \partial x_{j'}} V_k + \frac{\partial x_k}{\partial x_{j'}} \partial_{i'} V_k - \frac{\partial^2 x_l}{\partial x_{j'} \partial x_{i'}} V_l - \frac{\partial x_l}{\partial x_{i'}} \partial_{j'} V_l \\ & = \frac{\partial x_k}{\partial x_{j'}} \partial_{i'} V_k - \frac{\partial x_l}{\partial x_{i'}} \partial_{j'} V_l \\ & = \frac{\partial x_k}{\partial x_{j'}} \frac{\partial x_l}{\partial x_{i'}} \partial_{l} V_k - \frac{\partial x_l}{\partial x_{i'}} \frac{\partial x_k}{\partial x_{j'}} \partial_{k} V_l \\ & = \frac{\partial x_k}{\partial x_{j'}} \frac{\partial x_l}{\partial x_{i'}} (\partial_l V_k - \partial_k V_l) \\ & = \frac{\partial x_k}{\partial x_{j'}} \frac{\partial x_l}{\partial x_{i'}} T_{kl}. \end{align} Then we see that the $T_{ij}$ form the components of a type $(0,2)$ tensor.

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Bah - that's what I said! :) –  Neal Jan 22 '13 at 15:47

Alternatively to Neal, you can approach this in a coordinate-free way using the Jacobian and some wedge products.

In particular, if $f(x) = x'$ defines a coordinate transformation, then $\underline f(a) = a \cdot \partial f$ is the Jacobian, and $\overline f(a)$ denotes the transpose. Covectors transform according to the transpose. That is,

$$V = \overline f(V'), \quad \nabla = \overline f(\nabla')$$

where the primes refer to objects in the new coordinates. If $T$ is a tensor, it must follow $T = \overline f(T')$. We know that $T = \nabla \wedge V$, so $T' = \nabla' \wedge V'$ if $T$ is a tensor. Let's check that:

$$T = \nabla \wedge V = \overline f(\nabla') \wedge \overline f(V') = \overline f(\nabla ' \wedge V') = \overline f(T')$$

If you can convert that approach to index notation, you deserve credit for the problem.

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Thank you, Murphid. This is an interesting way! –  Nathaniel Jan 22 '13 at 15:40
    
+1 for your last sentence. –  Neal Jan 22 '13 at 15:46

Here are two paths you can take toward solving the problem.

  • You can take your definition of "tensor" as a set of indexed quantities which transform according to a certain rule. You need to show that $T_{ij} = \nabla_iT_j - \nabla_jT_i$ transforms according to these rules. You know the expression for the covariant derivative in terms of derivatives and Christoffel symbols, so expand out $\nabla_i T_j - \nabla_j T_i$ and simplify until you get a linear combination of quantities you already know are tensors. Then you're done, since a linear combination of tensors is again a tensor.

  • To work from a coordinate-free vantage point, you can use a slightly different definition of "tensor": a $(p,q)$-tensor is a pointwise linear map which takes as input $p$ vector fields and $q$ covector fields and returns a function. From this perspective, you have defined the function $T(X,Y) = \nabla_XY - \nabla_YX$. You would like to show it is a tensor. So show that it is pointwise linear in both inputs, i.e., that for any smooth functions $f,g$, $T(fX,gY) = fgT(X,Y)$.

For more, you might check out the Wikipedia page on the torsion tensor.

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Thank you, Neal! –  Nathaniel Jan 22 '13 at 15:41

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