Tensor transformations

We have a \(n \in \mathbb{N}\) finite dimensional vector space \(V\) such that \(\dim V = n\), with a basis \(\{\mathbf{e}_i\}_{i=1}^n,\) a corresponding dual space \(V^*\) with a basis \(\{\mathbf{\hat e}^i\}_{i=1}^n\) and a pseudo inner product \(\bm{g}\) on \(V.\)

Let us introduce a different basis \(\{\mathbf{f}_i\}_{i=1}^n\) of \(V\) with a corresponding dual basis \(\{\mathbf{\hat f}^i\}_{i=1}^n\) of \(V^*\) which are related to the former basis \(\{\mathbf{e}_i\}_{i=1}^n\) by

\[ \mathbf{f}_j = A^i_j \mathbf{e}_i, \]

so that \(\mathbf{\hat e}^i = A^i_j \mathbf{\hat f}^j\).

Transformation of tensors

Recall from the section of tensor-formalism that a holor depends on the chosen basis, but the corresponding tensor itself does not. This implies that holors transform in a particular way under a change of basis, which is characteristic for tensors.

Theorem 1: let \(\mathbf{T} \in \mathscr{T}^p_q(V)\) be a tensor with \(p=q=1\) without loss of generality and \(B = A^{-1}\). Then \(\mathbf{T}\) may be decomposed into

\[ \begin{align*} \mathbf{T} &= T^i_j \mathbf{e}_i \otimes \mathbf{\hat e}^j, \\ &= \overline T^i_j \mathbf{f}_i \otimes \mathbf{\hat f}^j, \end{align*} \]

with the holors related by

\[ \overline T^i_j = B^i_k A^l_j T^k_l. \]

Proof:

We have

\[ \begin{align*} \mathbf{T} &= T^i_j B^k_i \mathbf{f}_k \otimes A^j_l \mathbf{\hat f}^l,\\ &= B^k_i A^j_l T^i_j \mathbf{f}_k \otimes \mathbf{\hat f}^l,\\ &= \overline T^i_j \mathbf{f}_i \otimes \mathbf{\hat f}^j. \end{align*} \]

The homogeneous nature of the tensor transformation implies that a holor equation of the form \(T^i_j = 0\) holds relative to any basis if it holds relative to a particular one.

Transformation of volume forms

Lemma 1: let \((V, \bm{\mu})\) be a vector space with an oriented volume form with

\[ \begin{align*} \bm{\mu} &= \mu_{i_1 \dots i_n} \mathbf{\hat e}^{i_1} \otimes \cdots \otimes \mathbf{\hat e}^{i_n}, \\ &= \overline \mu_{i_1 \dots i_n} \mathbf{\hat f}^{i_1} \otimes \cdots \otimes \mathbf{\hat f}^{i_n}, \end{align*} \]

then we have

\[ \overline \mu_{j_1 \dots j_n} = A^{i_1}_{j_1} \cdots A^{i_n}_{j_n} \mu_{i_1 \dots i_n} = \mu_{j_1 \dots j_n} \det (A). \]

Proof:

We have

\[ \begin{align*} \bm{\mu} &= \mu_{i_1\dots i_n} A^{i_1}_{j_1} \mathbf{\hat f}^{j_1} \otimes \dots \otimes A^{i_n}_{j_n} \mathbf{\hat f}^{j_n},\\ &= A^{i_1}_{j_1} \cdots A^{i_n}_{j_n} \mu_{i_1\dots i_n} \mathbf{\hat f}^{j_1} \otimes \dots \otimes \mathbf{\hat f}^{j_n},\\ &= A^{i_1}_{j_1} \cdots A^{i_n}_{j_n} [i_1,\dots,i_n] \mathbf{\hat f}^{j_1} \otimes \dots \otimes \mathbf{\hat f}^{j_n},\\ &= \det(A) [j_1,\dots,j_n] \mathbf{\hat f}^{j_1} \otimes \dots \otimes \mathbf{\hat f}^{j_n},\\ &= \overline \mu_{i_1 \dots i_n} \mathbf{\hat f}^{i_1} \otimes \dots \otimes \mathbf{\hat f}^{i_1}. \end{align*} \]

Then \(\det(A)\) is the volume scaling factor of the transformation with \(A\). So that if \(\bm{\mu}(\mathbf{e}_1, \dots, \mathbf{e}_n) = 1\), then \(\bm{\mu}(\mathbf{f}_1, \dots, \mathbf{f}_n) = \det(A).\)

Theorem 2: let \((V, \bm{\mu})\) be a vector space with an oriented volume form with

\[ \begin{align*} \bm{\mu} &= \mu_{i_1 \dots i_n} \mathbf{\hat e}^{i_1} \otimes \cdots \otimes \mathbf{\hat e}^{i_n}, \\ &= \overline \mu_{i_1 \dots i_n} \mathbf{\hat f}^{i_1} \otimes \cdots \otimes \mathbf{\hat f}^{i_n}, \end{align*} \]

and if we define

\[ \overline \mu_{i_1 \dots i_n} \overset{\text{def}}{=} \frac{1}{\det (A)} A^{j_1}_{i_1} \cdots A^{j_n}_{i_n} \mu_{j_1 \dots j_n}, \]

then \(\mu_{i_1 \dots i_n} = \overline \mu_{i_1 \dots i_n} = [i_1, \dots, i_n]\) is an invariant holor.

Transformation of Levi-Civita form

Theorem 3: let \(\bm{\epsilon} \in \bigwedge_n(V)\) be the Levi-Civita tensor with

\[ \begin{align*} \bm{\epsilon} &= \epsilon_{i_1 \dots i_n} \mathbf{\hat e}^{i_1} \otimes \cdots \otimes \mathbf{\hat e}^{i_n}, \\ &= \overline \epsilon_{i_1 \dots i_n} \mathbf{\hat f}^{i_1} \otimes \cdots \otimes \mathbf{\hat f}^{i_n}, \end{align*} \]

then \(\epsilon_{i_1 \dots i_n} = \overline \epsilon_{i_1 \dots i_n}\) is an invariant holor.

Proof:

Follows directly from the definition \(\bm{\epsilon} = \sqrt{g} \bm{\mu}\).