I'm studying general relativity and tensors. It seems that in the cooridnate independent form of the tensor, the order of indices matters even between an upper and lower index. For example, in general,

$$T^{\ \ \nu}_{\mu} \neq T^{\nu}_{\ \ \mu}$$

I can see this because when trying to lower the $\nu$ index using the metric $g_{\mu\nu}$, the lower index occurs in a different position in the final answer. However, my question is that do both $T^{\ \ \nu}_{\mu}$ and $T^{\nu}_{\ \ \mu}$ belong to the same tangent space at a point in the manifold. If so, they can be represented using the same basis set. Then in the matrix formulation, how is $T^{\ \ \nu}_{\mu}$ different from $T^{\nu}_{\ \ \mu}$, given that the upper index refers to the row and the lower index refers to the column (I think this is the usual convention?).


1 Answer 1


They don't belong to the same tangent space (in fact, they don't belong to any tangent space) and cannot be represented using the same basis. The abstract tensor with components $T^{\ \ \nu}_{\mu}$ lives in the tensor product $V^* \otimes V$ and the abstract tensor with components $T^{\nu}_{\ \ \mu}$ is lives in the tensor product $V \otimes V^*$, where $V$ is the tangent space of the manifold at some point $x$ and $V^*$ is the dual space of $V$. A basis for the first is $\{\mathbf{\varepsilon}^\mu \otimes \mathbf{e}_\nu\}$ and a basis for the second is $\{\mathbf{e}_\mu \otimes \mathbf{\varepsilon}^\nu\}$ where $\{\mathbf{e}_\mu\}$ is a basis for $V$ and $\{\mathbf{\varepsilon}^\mu\}$ is a basis for $V^*$.

Tensors can have any number of indices. But even when they have only two, you shouldn't think of the upper index as the row and the lower index as the column. You should think of the first index as the row and the second index as the column, because this is the convention generally used when displaying tensors of rank 2 as matrices.

As matrices, $T^{\ \ \nu}_{\mu}$ and $T^{\mu}_{\ \ \nu}$ are different because to lower or raise an index you contract with the metric tensor or its inverse. When you lower or raise an index, it stays in the same position (i.e., first, second, etc.)

For example, in Minkowski space with signature (+---), the contravariant components of the electromagnetic field tensor are $$F^{\mu\nu}=\begin{bmatrix} 0 && -E_x && -E_y && -E_z \\ E_x && 0 && -B_z && B_y \\ E_y && B_z && 0 && -B_x \\ E_z && -B_y && B_x && 0 \\ \end{bmatrix},$$

the covariant components are $$F_{\mu\nu}=\begin{bmatrix} 0 && E_x && E_y && E_z \\ -E_x && 0 && -B_z && B_y \\ -E_y && B_z && 0 && -B_x \\ -E_z && -B_y && B_x && 0 \\ \end{bmatrix},$$

and the two mixed forms are $$F^{\mu}_{\ \ \ \nu}=\begin{bmatrix} 0 && E_x && E_y && E_z \\ E_x && 0 && B_z && -B_y \\ E_y && -B_z && 0 && B_x \\ E_z && B_y && -B_x && 0 \\ \end{bmatrix}$$

and $$F^{\ \ \nu}_{\mu}=\begin{bmatrix} 0 && -E_x && -E_y && -E_z \\ -E_x && 0 && B_z && -B_y \\ -E_y && -B_z && 0 && B_x \\ -E_z && B_y && -B_x && 0 \\ \end{bmatrix}.$$

(I hope I got all the signs right.)

  • $\begingroup$ I haven't worked it out, but I imagine that for a symmetric tensor, the components of $T^\mu{}_\nu$ and $T_\mu{}^\nu$ are the same...? $\endgroup$
    – user4552
    Nov 16, 2018 at 1:38
  • $\begingroup$ Yes. Suppose $T_{\mu\nu}=T_{\nu\mu}$. Then $$T^\mu_{\;\;\nu}=g^{\mu\lambda}T_{\lambda\nu}=g^{\mu\lambda}T_{\nu\lambda}=T_\nu^{\;\;\mu}.$$ A symmetric 2-index tensor is symmetric regardless of whether the indices are up, down, or mixed. $\endgroup$
    – G. Smith
    Nov 16, 2018 at 4:50
  • $\begingroup$ The same applies to antisymmetry: Antisymmetric contravariant components $F^{\mu\nu}=-F^{\nu\mu}$ imply antisymmetric covariant components $F_{\mu\nu}=-F_{\nu\mu}$ and antisymmetric mixed components $F^\mu_{\;\;\nu}=-F_\nu^{\;\;\mu}$. The first two mean that the matrices representing $F{\mu\nu}$ and $F{\mu\nu}$ are antisymmetric, but in the mixed case the matrices are NOT antisymmetric! The antisymmetric tensor relation is a relation between two different matrices (the third and the fourth above) in this case. $\endgroup$
    – G. Smith
    Nov 16, 2018 at 5:41
  • $\begingroup$ I suppose this can happen in the symmetric case as well: $T^\mu_{\;\;\nu}$ may not be a symmetric matrix, but it is symmetrically related to $T_\nu^{\;\mu}$. $\endgroup$
    – G. Smith
    Nov 16, 2018 at 5:45

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