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 $$\frac{\partial \phi}{\partial x^\alpha}:=\phi_{,\alpha}.$$