^^ [[Schutz - 2 - Differentiable Manifolds and Tensors|2. Differentiable Manifolds and Tensors]] << [[Schutz - 2.25 - Contraction|2.25 Contraction]] | [[Schutz - 2.27 - Tensor Operations on Components|2.27 Tensor Operations on Components]] >> # Basis Transformations In older literature, tensors were defined by their transformation rules under a change of basis. In this section, we consider [[Schutz - 2.7 - Vectors and Vector Fields#^ab2ebc|vectors]] and [[Schutz - 2.22 - Tensors and Tensor Fields#^472d5f|tensors]] defined on some point $P \in M$. Suppose we begin with a vector basis $\{\pmb{e}_i\}$ and wish to transform to a new basis $\{\pmb{e}_{j'}\}$. Primes on the indices denote a new basis. Then, in the [[Schutz - 2.7 - Vectors and Vector Fields#^4e794a|tangent space]] $T^p$ there is a linear transformation $\Lambda$ from the old basis to the new: $ \pmb{e}_{j'} = \Lambda^i_{\,j'} \pmb{e}_i $ The matrix $\Lambda^i_{\,j'}$ is non-singular to guarantee linear independence of the new-basis, but otherwise arbitrary. It is not the collection of components of a tensor since the indices refer to two different bases. Moreover, the old one-form satisfies: $ \tilde{\omega}^i(\pmb{e}_k) = \delta^i_{\, k} $ Thus, we get: $ \tilde{\omega}^i(\pmb{e}_{j'}) = \tilde{\omega}^i(\Lambda^i_{\,j'} \pmb{e}_k) = \Lambda^k_{\,j'}\delta^i_{\, k} = \Lambda^i_{\,j'} $ We define the inverse of $\Lambda^i_{\,\,j'}$ to be $\Lambda^{k'}_{\,\,j}$, giving: $ \begin{align} \Lambda^{k'}_{\,\,i}\Lambda^i_{\,\,j'} &= \delta^{k'}_{\,j'}\\ \Lambda^{k'}_{\,\,j}\Lambda^i_{\,\,k'} &= \delta^{i}_{\,j} \end{align} $ Multiply above by $\Lambda^{k'}_{\,\,i}$: $ \Lambda^{k'}_{\,\,i}\tilde{\omega}^i(\pmb{e}_{j'}) = \Lambda^{k'}_{\,\,i}\Lambda^i_{\,\,j'} = \delta^{k'}_{\,\,j'} = \tilde{\omega}^{k'}(\pmb{e}_{j'}) $ Thus we get the transformation law: $ \tilde{\omega}^{j'} = \Lambda^{j'}_{\,\,i}\tilde{\omega}^i $ And we can now transform components simply as (using Linearity and identities defining the components): $ \begin{align} V^{i'} &= \tilde{\omega}^{i'}(\pmb{V}) = \Lambda^{i'}_{\,\,j} \tilde{\omega}^{i}(\pmb{V}) = \Lambda^{i'}_{\,\,j} V^j, \\ q_{i'} &= \tilde{q}(\pmb{e}_{i'}) = \Lambda^{j}_{\,\,i'} \tilde{q}(\pmb{e}_j) = \Lambda^{j}_{\,\,i'} q_j \end{align} $ And similarly for tensors of higher type. > **Transformation Laws for Tensors** > Bases transform as: >$ \begin{align} \pmb{e}_{j'} &= \Lambda^i_{\,j'} \pmb{e}_i, \\ \tilde{\omega}^{j'} &=\Lambda^{j'}_{\,\,i}\tilde{\omega}^i, \end{align} > $ > While components transform the opposite way: > $ \begin{align} V^{i'} &= \Lambda^{i'}_{\,\,j} V^j, \\ q_{i'} &= \Lambda^{j}_{\,\,i'} q_j \end{align} >$ > Tensors transform similarly. For an $(N, N')$ tensor: >$ T^{i_{1}' \ldots i_N'}_{\quad \quad \,\,\,j_{1}'\ldots j_{N'}'} = \Lambda^{i'_1}_{\,\,k_1} \ldots \Lambda^{i'_N}_{\,\,k_N} \Lambda^{l_1}_{\,\,j'_1} \ldots \Lambda^{l_N}_{\,\,j'_{N'}} T^{k_{1} \ldots k_N}_{\quad \quad \,\,\,l_{1}\ldots l_{N'}} >$ > **Covectors and Contravectors** > In the old terminology, one-forms are called *covectors* because they transform like the basis does, while vectors are called contravectors because they transform contra (i.e. opposite) to the basis. "co-goes-low" is a helpful mnemonic to remember where the indices go. ## Coordinate Transformations Consider a region of a manifold $U \subset M$ that has a coordinate system $\{x^i\}$ and we wish to introduce new functions $\{y^{i'}\}$ given by the equations: $ y^{i'} = f^{i'}(x^j), i', \quad j = 1, \ldots n $ These equations are a coordinate transformation if the Jacobian matrix of partial derivatives $\partial y^{i'}/ \partial x^j$ has a nonzero determinant in $U$ (to guarantee invertability, and this is also equivalent to saying that the $\{y^{i'}\}$ are linearly independent). A given point $P \in U$, can be described by two different sets of numbers, $\{x^i\}$ or $\{y^{i'}\}$. Likewise, at $P$ we have two different [[Schutz - 2.8 - Basis Vectors and Basis Vector Fields#^cf2d81|coordinate vector bases]], $\{\partial /\partial x^i\}$ and $\{\partial /\partial y^{i'}\}$. By the chain rule: $ \frac{\partial}{\partial y^{i'}} = \frac{\partial x^i}{\partial y^{i'}} \frac{\partial}{\partial x^{i}} $ Thus we conclude: $ \Lambda^{i}_{\,\,i'} = \frac{\partial x^{i}}{\partial y^{i'}} $ With the inverse matrix: $ \Lambda^{i'}_{\,\,i} = \frac{\partial y^{i'}}{\partial x^{i}} $ as one can see from: $ \frac{\partial x^{i}}{\partial y^{i'}} \frac{\partial y^{i'}}{\partial y^{k}} = \delta^{i}_{\,k} $