The proof of the tensor character of Τ, follows immediately from the expressions (8), (10) or (12), or the transformation equations (9), (11), (13); equations (8), (10) and (12) are themselves examples of the outer multiplication of tensors of the first rank. _Reduction in rank of a mixed Tensor._ From every mixed tensor we can get a tensor which is two ranks lower, when we put an index of co-variant character equal to an index of the contravariant character and sum according to these indices (Reduction). We get for example, out of the mixed tensor of the fourth rank A_{αβ}^{γδ}, the mixed tensor of the second rank A_{β}^{δ} = A_{αβ}^{αδ} = (∑_{α} A_{αβ}^{αδ}) and from it again by “reduction” the tensor of the zero rank A = A_{β}^{β} = A_{αβ}^{αβ}. The proof that the result of reduction retains a truly tensorial character, follows either from the representation of tensor according to the generalisation of (12) in combination with (6) or out of the generalisation of (13). _Inner and mixed multiplication of Tensors._ This consists in the combination of outer multiplication with reduction. Examples:—From the co-variant tensor of the second rank A_{μν} and the contravariant tensor of the first rank B^{σ} we get by outer multiplication the mixed tensor D^{σ}_{μν} = A_{μν} B^{σ} . Through reduction according to indices ν and σ (_i.e._, putting ν = σ), the co-variant four vector D_{μ} = D^{ν}_{μν} = A_{μν} B^{ν} is generated. These we denote as the inner product of the tensor A_{μν} and B^{σ}. Similarly we get from the tensors A_{μν} and B^{στ} through outer multiplication and two-fold reduction the inner product A_{μν} B^{μν}. Through outer multiplication and one-fold reduction we get out of A_{μν} and B^{στ}, the mixed tensor of the second rank D^{τ}_{μ} = A_{μν} B^{τν}. We can fitly call this operation a mixed one; for it is outer with reference to the indices μ and τ and inner with respect to the indices ν and σ. We now prove a law, which will be often applicable for proving the tensor-character of certain quantities. According to the above representation, A_{μν} B^{μν} is a scalar, when A_{μν} and B^{στ} are tensors. We also remark that when A_{μν} B^{μν} is an invariant for every choice of the tensor B^{μν}, then A_{μν} has a tensorial character. Proof:—According to the above assumption, for any substitution we have A_{στ′} B^{στ′} = A_{μν} B^{μν}. From the inversion of (9) we have however $$ B_{\mu \nu} = \frac{\partial x_{\mu}}{\partial x_{\sigma'}} \frac{\partial x_{\nu}}{\partial \tau'} B^{\sigma \tau'} $$ Substitution of this for B^{μν} in the above equation gives $$ (A_{\sigma \tau'} - \frac{\partial x_{\mu}}{\partial x_{\sigma'}} \frac{\partial x_{\nu}}{\partial x_{\tau'}}) B^{\sigma \tau'} = 0 $$