Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I would like to ask something

On the Barrett Oneill's Semi-Riemann Geometry there is a definition of tensor component: Let $\xi=(x^1,\dots ,x^n)$ be a coordinate system on $\upsilon\subset M$. If $A \in \mathscr I^r_s (M)$ the components of $A$ relative to $\xi$ are the real-valued functions $A _j^i, \dots j^i =A(dx^i_1,\dots ,dx^i_s, \delta_{j 1},\dots, \delta_{j s})$ (I couldn't write it properly but $i$'s are from $1$ to $r$ and $j$'s are $1$ to $s$) on $\upsilon$ where all indices run from $1$ to $n=dim M$.

(and here is the thing that I didn't get)
By the definition above the $i$th component of $X$ relative to $\xi$ is $X(dx^i)$,which is interpreted as $dx^i(X)=X(x^i)$.

While one-forms are $(0,1)$ tensors we could interprete them like $V(\theta)=\theta(V)$
So we can do the same thing here: $X(dx^i)=dx^i(X)$.
But how did we write $dx^i(X)=X(x^i)$

Do I think something wrong?
(Sorry if I wrote it bad but I couldn't find how to write.)

share|improve this question
add comment

2 Answers

The is a very nice intuitive explanation of this in Penrose's Road to Reality, ch 14. As a quick summary, any vector field can be though of as a directional derivative operator on scalar valued functions, i.e given a scalar valued function $f$ and vector field $X$, $X(f)$ is a linear approximation the change in $f$ when displaced by $X$ at every point. Given a coordinate system $x^i$, this can be formalized by defining $X \doteq \Sigma_i X^i \partial_{x^i}$ which is a linear operator, with the $\partial_{x^i}$ acting as a basis for the space of vector fields. 1-Forms are co-vectors, so they must be linear maps from vectors to reals, with $dx^i$ as the dual basis. For exact 1-forms this is defined as $df(X) \doteq X(f) = \Sigma_i X^i \partial_{x^i}f$ which formally restates the directional derivative picture. For the $dx^i$ this means $dx^i(\partial_{x^j}) = \partial_{x^j} x^i = \delta^i_j$ and $dx^i(X) = X(x^i) = \Sigma_j X^j \partial_{x^j} x^i = X^i$ as required for a dual basis.

To formalize this for the general Riemannian setting, you need express it in terms of manifolds, charts and tangent spaces/bundles, with appropriate smoothness conditions on everything, but Penrose's explanation gives you a nice mental picture to start with. The book also has excellent diagrams. It is worth getting just for differential geometry chapter.

share|improve this answer
add comment

The tensor $A_p$ at $p\in M$ of type $(r,s)$ is the multilinear map

$$A_p: \underbrace{T_p^{*} M\times\dots T_p^{*} M}_{r~\text{times}} \times\underbrace{T_p M\times\dots T_p M}_{s~\text{times}}\rightarrow \mathbb R $$

whose components ${A_p}^{i_1\dots i_r}_{j_1\dots j_s}$ are given by

$${A_p}^{i_1\dots i_r}_{j_1\dots j_s}:=A_p(dx^{i_1},\dots,dx^{i_r},\frac{\partial}{\partial x_{j_1}},\dots, \frac{\partial}{\partial x_{j_s}}),$$

denoting by $\xi=(x_1,\dots,x_n)$ a local coordinate system at $p$, and by $\{dx^{\bullet}\}$, $\{\frac{\partial}{\partial x_{\bullet}}\}$ dual basis of $T_p^{*} M$, $T_p M$.

As stated,

$$A_p^{i}:=A_p(dx^i), $$ $${A_p}_j:=A_p(\frac{\partial}{\partial x_j}). $$

The above convenstions and definitions are those which are currently used in most textbooks. I would propose then to stop here and use them for your computations; what comes under "which is interpreted...", is in my opinion not clear (for example, what is $A(x_i)?$ Maybe $A$ evaluated at $(x_1,\dots,x_n)$? But then why to use that notation, if it is what has been using since the beginning of the OP? Etc...).

share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.