General Relativity/Rigorous Definition of Tensors

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We have seen that a 1-form ("covariant vector") can be thought of an operator with one slot in which we insert a vector ("contravariant vector") and get the scalar \mathbf{\sigma} \left( \mathbf{v} \right). Similarly, a vector can be thought of as an operator with one slot in which we can insert a 1-form to obtain the scalar \mathbf{v} \left( \mathbf{\sigma} \right). As operators, they are linear, i.e., \mathbf{\sigma} \left( \alpha \mathbf{u} +\beta \mathbf{v} \right) = \alpha \mathbf{\sigma} \left( \mathbf{u} \right) + \beta \mathbf{\sigma} \left( \mathbf{v} \right).

A tensor of rank n is an operator with n slots for inserting vectors or 1-forms, which, when all n slots are filled, returns a scalar. In order for such an operator to be a tensor, it must be linear in each slot and obey certain transformation rules (more on this later). An example of a rank 2 tensor is \mathbf{T} = T^\mu_{\ \nu} \mathbf{e}_\mu \otimes \mathbf{d}x^\nu. The symbol \otimes (pronounced "tensor") tells you which slot each index acts on. This tensor \mathbf{T} is said to be of type (1,1) because it has one contravariant slot and one covariant slot. Since \mathbf{e}_\mu acts on the first slot and \mathbf{d}x^\nu acts on the second slot, we must insert a 1-form in the first slot and a vector in the second slot (remember, 1-forms act on vectors and vice-versa). Filling both of these slots, say with \mathbf{\sigma} and \mathbf{u}, will return the scalar \mathbf{T} \left( \mathbf{\sigma}, \mathbf{u} \right). We can use linearity (remember, the tensor is linear in each slot) to evaluate this number:

\mathbf{T} \left( \mathbf{\sigma}, \mathbf{u} \right)
= T^\mu_{\ \nu} \mathbf{e}_\mu \otimes \mathbf{d}x^\nu \left( \sigma_\alpha \mathbf{d}x^\alpha, u^\beta \mathbf{e}_\beta \right) 
= T^\mu_{\ \nu} \mathbf{e}_\mu \left( \sigma_\alpha \mathbf{d}x^\alpha \right) \mathbf{d}x^\nu \left( u^\beta \mathbf{e}_\beta \right)
= T^\mu_{\ \nu} \sigma_\alpha u^\beta \delta^\alpha_\mu \delta^\nu_\beta
= T^\mu_{\ \nu} \sigma_\mu u^\nu

We don't have to fill all of the slots. This will of course not produce a scalar, but it will lower the rank of the tensor. For example, if we fill the second slot of \mathbf{T}, but not the first, we get a rank 1 tensor of type (1,0) (which is a contravariant vector):

\mathbf{T} \left( \cdot \ , \mathbf{u} \right) 
= T^\mu_{\ \nu} \mathbf{e}_\mu \otimes \mathbf{d}x^\nu \left( \cdot \ , u^\gamma \mathbf{e}_\gamma \right) 
= T^\mu_{\ \nu} \mathbf{e}_\mu \left( \cdot \right) \mathbf{d}x^\nu \left( u^\gamma \mathbf{e}_\gamma \right) 
=T^\mu_{\ \nu} \mathbf{e}_\mu u^\gamma \delta^\nu_\gamma = T^\mu_{\ \nu} u^\nu \mathbf{e}_\mu

For another example, consider the rank 5 tensor \mathbf{S}=S^{\alpha \ \ \mu \nu} _{\ \beta \gamma} \mathbf{e}_\alpha \otimes \mathbf{d}x^\beta \otimes \mathbf{d}x^\gamma \otimes \mathbf{e}_\mu \otimes \mathbf{e}_\nu. This is a tensor of type (3,2). We can fill all of its slots to get a scalar:

\mathbf{S} \left( \mathbf{\sigma}, \mathbf{u}, \mathbf{v}, \mathbf{\rho}, \mathbf{\xi} \right)
= S^{\alpha \ \ \mu \nu} _{\ \beta \gamma} \sigma_\alpha u^\beta v^\gamma \rho_\mu \xi_\nu

Filling only the 3rd and 4th slots, we get a rank 3 tensor of type (2,1):

\mathbf{S} \left( \cdot \ , \cdot \ , \mathbf{v}, \mathbf{\rho}, \cdot \right)
= S^{\alpha \ \ \mu \nu} _{\ \beta \gamma} v^\gamma \rho_\mu \mathbf{e}_\alpha \otimes \mathbf{d}x^\beta \otimes \mathbf{e}_\nu

As a final note, it should be mentioned that in General Relativity we will always have a special tensor called the "metric tensor" which will allow us to convert contravariant indices to covariant indices and vice-versa. This way, we can change the tensor type (n,m) and be able to insert either 1-forms or vectors into any slot of a given tensor.