tensor transformations

The present entry employs the terminology and notation definedand described in the entry on tensor arrays and basic tensors. To keep thingsreasonably self contained we mention that the symbol ${\mathrm{T}}^{p,q}$ refersto the vector space of type $(p,q)$ tensor arrays, i.e. maps

$$I^p\times I^q\to 𝕂,$$

where $I$ is some finite list ofindex labels, and where $𝕂$ is a field. The symbols ${\epsilon }_{(i)},{\epsilon }^{(i)},i\in I$ refer to the column and row vectors giving thenatural basis of ${\mathrm{T}}^{1,0}$ and ${\mathrm{T}}^{0,1}$, respectively.

Let $I$ and $J$ be two finite lists of equal cardinality, and let

$$T:{𝕂}^I\to {𝕂}^J$$

be a linear isomorphism. Everysuch isomorphism is uniquely represented by an invertible matrix

$$M:J\times I\to 𝕂$$

with entries given by

$$M_i^j={\left(T{\epsilon }_{(i)}\right)}^j,i\in I,j\in J.$$

In other words, theaction of $T$ is described by the following substitutions

$${\epsilon }_{(i)}\mapsto \sum _{j\in J}{M}_i^j{\epsilon }_{(j)},i\in I.$$(1)

Equivalently, the action of $T$ is given by matrix-multiplicationof column vectors in ${𝕂}^I$ by $M$.

The corresponding substitutions relations for the type $(0,1)$ tensorsinvolve the inverse matrix $M^{-1}:I\times J\to 𝕂,$ andtake the form

$${\epsilon }^{(i)}\mapsto \sum _{j\in J}{\left(M^{-1}\right)}_j^i{\epsilon }^{(j)},i\in I.$$(2)

The rules for type $(0,1)$ substitutions are what they are, because ofthe requirement that the ${\epsilon }_{(i)}$ and ${\epsilon }^{(i)}$ remain dual baseseven after the substitution. In other words we want the substitutionsto preserve the relations

$${\epsilon }^{(i_1)}{\epsilon }_{(i_2)}={\delta }_{i_2}^{i_1},i_1,i_2\in I,$$

where the left-hand side of the above equation features the innerproduct and the right-hand side the Kronecker delta. Given that thevector basis transforms as in (1) and given the aboveconstraint, the substitution rules for the linear form basis, shown in(2), are the only such possible.

The classical terminology of contravariant and covariant indices ismotivated by thinking in term of substitutions. Thus, suppose weperform a linear substitution and change a vector, i.e. a type $(1,0)$tensor, $X\in {𝕂}^I$ into a vector $Y\in {𝕂}^J$. The indexedvalues of the former and of the latter are related by

$$Y^j=\sum _{i\in I}{M}_i^j{X}^i,j\in J.$$(3)

Thus, we seethat the “transformation rule” for indices is contravariant to thesubstitution rule (1) for basis vectors.

In modern terms, this contravariance is best described by saying thatthe dual space space construction is a contravariant functor. In otherwords, the substitution rule for the linear forms, i.e. the type$(0,1)$ tensors, is contravariant to the substitution rule forvectors:

$${\epsilon }^{(j)}\mapsto \sum _{i\in I}{M}_i^j{\epsilon }^{(i)},j\in J,$$(4)

in full agreement with the relation shown in (2).Everything comes together, and equations (3) and(4) are seen to be one and the same, once we remarkthat tensor array values can be obtained by contracting withcharacteristic arrays. For example,

$$X^i={\epsilon }^{(i)}(X),i\in I;Y^j={\epsilon }^{(j)}(Y),j\in J.$$

Finally we must remark that the transformation rule for covariantindices involves the inverse matrix $M^{-1}$. Thus if $\alpha \in {\mathrm{T}}^{0,1}(I)$ is transformed to a $\beta \in {\mathrm{T}}^{0,1}$ theindices will be related by

$${\beta }_j=\sum _{i\in I}{\left(M^{-1}\right)}_j^i{\alpha }_i,j\in J.$$

The most general transformation rule for tensor array indices istherefore the following: the indexed values of a tensor array $X\in {\mathrm{T}}^{p,q}(I)$ and the values of the transformed tensor array$Y\in {\mathrm{T}}^{p,q}(J)$ are related by

$$Y_{l_1\mathrm{\dots }{l}_q}^{j_1\mathrm{\dots }{j}_p}=\sum _{\genfrac{}{}{0.0pt}{}{i_1,\mathrm{\dots },i_p\in I^p}{k_1,\mathrm{\dots },k_q\in I^q}}{M}_{i_1}^{j_1}\mathrm{\cdots }{M}_{i_p}^{j_p}{\left(M^{-1}\right)}_{l_1}^{k_1}\mathrm{\cdots }{\left(M^{-1}\right)}_{l_q}^{k_q}{X}_{k_1\mathrm{\dots }{k}_q}^{i_1\mathrm{\dots }{i}_p},$$

for all possible choice of indices$j_1,\mathrm{\dots }{j}_p,l_1,\mathrm{\dots },l_q\in J.$Debauche of indices, indeed!