Vector, One-form and Tensor

Vectors

Vectors

The components of a vector is denoted by $$p^\alpha,$$ where \(\alpha\) indicates \(\alpha\)th component of \(\vec{p}\). For example, for a 2-vector, \(\alpha\) can be 0, 1. \(p^{0}\) is the 0th component of \(\vec{p}\) and \(p^{1}\) is the 1st component of \(\vec{p}\).

Basis

One-form

One-form is a function that takes a vector and output a scalar $$\tilde{p}:\vec{A}\to \mathbb{R}.$$

One-forms satisfy certain conditions \begin{align} \tilde{s} &= \tilde{p} + \tilde{q} \\ \tilde{r} &= \alpha \tilde{p}, \end{align} where \(\tilde{s}\), \(\tilde{q}\) and \(\tilde{r}\) are one-forms while \(\alpha\) is a scalar, such that one-forms can form a vector space, called dual vector space.

The components of a one-form is denoted by $$p_\alpha,$$ where \(\alpha\) indicates \(\alpha\)th component of \(\tilde{p}\), just like vectors. Upper index indicates the component of a vector while lower index indicates the component of a one-form.

The operation of a one-form \(\tilde{p}\) on a vector \(\vec{A}\) $$\tilde{p}(\vec{A}) = p_\alpha A^{\alpha}$$ is called contraction. Here, Einstein summation is used $$p_\alpha A^\alpha = \sum^N_{\alpha=0}p_\alpha A^\alpha,$$ where \(N\) is the dimension of the vector space.

A set of linear independent one-forms that span the dual vector space can also be found. Let the set be denoted as \(\{\tilde{\omega}^\alpha\}\). For an arbitrary one-form \(\tilde{p}\), \begin{align} \tilde{p} &= p_\alpha\tilde{\omega}^\alpha\\ \tilde{p}(\vec{A})&= p_\alpha A^{\alpha}\\ &= p_{\alpha}\tilde{\omega}^{\alpha}(\vec{A})\\ &= p_{\alpha}\tilde{\omega}^{\alpha}(A^\beta\vec{e}_\beta)\\ &= p_{\alpha}A^{\beta}\tilde{\omega}^{\alpha}(\vec{e}_\beta)\\ &= p_\alpha A^\beta \tilde{\omega}^\alpha(\vec{e}_\beta). \end{align} Then, we conclude that $$ \bbox[5px,border:2px solid #666] { \tilde{\omega}^{\alpha}(\vec{e}_\beta) = \delta^\alpha_{\beta}, } $$ that is to say, when we have a set of basis \(\{\vec{e}_\alpha\}\), a set of \(\{\tilde{\omega}^\alpha\}\) can also be found.

Tensor

$$\begin{pmatrix} M \\ N \end{pmatrix} $$ Components \(\bf{f}\) a \begin{pmatrix}0\\ 2\end{pmatrix} tensor $$f_{\alpha\beta}:=\bf{f}(\vec{e}_\alpha,\vec{e}_beta)$$

Basis. Let \(\tilde{\omega}^{\alpha\beta}\) be the basis of \(\bf{f}\) $$\bf{f} &= f_{\alpha\beta}\tilde{\omega}^{\alpha\beta}$$ \begin{align} f_{\mu\nu} &= \bf{f}(\vec{e}_\mu,\vec{e}_\nu)\\ &= f_{\alpha\beta}\tilde{\omega}^{\alpha\beta}(\vec{e}_\mu,\vec{e}_\nu). \end{align} Since \begin{align} \tilde{\omega}^{\alpha\beta}(\vec{e}_\mu,\vec{e}_\nu) &= \delta^\alpha_\mu\delta^\beta_\nu\\ \tilde{\omega}^{\alpha\beta} &= \tilde{\omega}^{\alpha}\otimes \tilde{\omega}^{\beta}, \end{align} $$\bf{f}=f_{\alpha\beta}\tilde{\omega}^\alpha\otimes\tilde{\omega}^\beta$$

Metric Tensor

Lowering index $$T^{\alpha}_{\beta\gamma} = \eta_{\beta\mu}T^{\alpha\mu}_{\gamma}$$ Raising index $$T^{\alpha\beta\gamma} = \eta^{\gamma\mu}T^{\alpha\beta}_\mu$$ $$\eta^\alpha_\beta\equiv \eta^{\alpha\mu}\eta_{\mu\beta} = \delta^\mu_\beta$$

Derivative Denotation

The derivative of \(\phi\) w.r.t \(x^{\alpha}\) is denoted by $$\frac{\partial \phi}{\partial x^\alpha}:=\phi_{,\alpha}.$$