Understanding basics of tensors

I am trying to understand tensors to learn General relativity. In the book that I am reading they claim that if the basis of a vector space undergoes a linear transformation $T$ then the components of the vector undergo the linear transformation $(T^{-1})^T$ (transpose of inverse of $T$). This is how they prove it.

If the initial basis of a vector space is $\{e_i\}_1^n$ and the final basis $\{f_i\}_1^n$ is related by $$f_i=T_i{}^je_j$$ then for an arbitrary vector $v=x^ie_i=y^jf_j$ we have, $$y^k=x^j(T^{-1})_j{}^k=((T^{-1})^T)^k{}_jx^j$$

I am not able to understand the last equality at all. I believe it is false. Can somebody clarify if it is correct or wrong?

• Related: physics.stackexchange.com/q/169762/2451 and links therein. Commented May 15, 2016 at 9:53
• Hi Coniferous, have you checked the link Qmechanic suggests? I believe it does answer your question but I'm reluctant to close this as a duplicate unless you agree. Commented May 15, 2016 at 10:15
• @JohnRennie I am not able to find answer to my question using the link suggested. What I want to understand is how $x^j(T^{-1})_j^k=((T^{-1})^T)_j^kx^j.$ If i use the concepts of the link given I think I get $x^j(T^{-1})_j^k=((T^{-1})^T)_k^j x^j.$ I am not sure if it is true. Commented May 15, 2016 at 10:30
• I've edited the question to make indices not be above each other. In the process I have in fact changed the order (in the LaTeX) of the indices of the last expression, which I am reasonably sure was your intent in fact. If it wasn't please revert the edit.
– user107153
Commented May 15, 2016 at 11:34

I'll use a notation where I write vectors as $\vec{v}$ (because that's how \vec renders and I have no idea how to turn it into boldface in MathJax) and use primes to indicate the different basis: $v'^i$ is the components of $\vec{v}$ in the primed basis (some people use primes on the indices, which I find confusing).

So given some basis $\left\{\vec{e}_i\right\}$ we can write any vector $\vec{v}$ as

$$\vec{v} = v^i\vec{e}_i$$

Now consider a new basis $\left\{\vec{e'}_i\right\}$ which is related to the original basis by some nonsingular transformation matrix $T$:

$$\vec{e'}_j = T_j{}^i\vec{e}_i$$

(Note that this is just matrix multiplication using the ESC, and in particular $T$ is not a tensor.)

Well, we can express $\vec{v}$ in the new basis:

\begin{align} \vec{v} &= v'^j\vec{e'}_j\\ &= v'^jT_j{}^i\vec{e}_i\\ &= v^i\vec{e}_i \end{align}

And comparing the second-last and last lines of this we get

$$v^i = v'^jT_j{}^i$$

OK, well we know $T$ is non-singular so there exists a $T^{-1}$, such that, in components

$$T_j{}^i\left(T^{-1}\right)_i{}^k = \delta_j{}^k$$

So let's multiply the expression for $v^i$ above by this on the right:

\begin{align} v^i\left(T^{-1}\right)_i{}^k &= v'^j T_j{}^i\left(T^{-1}\right)_i{}^k\\ &= v'^j\delta_j{}^k\\ &= v'^k \end{align}

or (big fanfare, and renaming indices gratuitously)

$$v'^i = v^j \left(T^{-1}\right)_j{}^i$$

And finally, we want to make this look like matrix multiplication on the left, so we need to diddle the indices of $T$:

$$v'^i = \left(\left(T^{-1}\right)^T\right)^i{}_j v^j$$

And we're done.

• "I have no idea how to turn it into boldface in MathJax"- \textbf{write here} is the way to write bold
– UKH
Commented May 15, 2016 at 11:21
• @Unnikrishnan: sorry, what I meant was I don't know how (in MathJax: I know in LaTeX) to make \vec{...} render as boldface.
– user107153
Commented May 15, 2016 at 11:27
• \vec{\textbf{write here}}
– UKH
Commented May 15, 2016 at 11:31
• What I want is $\mathbf{v}$ instead of $\vec{b}: I don't want the annoying arrow at all (but I want to write \vec). In LaTex I'd write \renewcommand*\vec[1]{\mathbf{#1}}. – user107153 Commented May 15, 2016 at 11:38 • @tfb one thing that I still donot understand is how you argue the last step. Let me use$S=T^{-1}$then what you did in last step is$v^jS_\mathbf{j}^\mathbf{i}=(S^T)_\mathbf{j}^\mathbf{i}v^j.$Notice that the indices for$S$didnot change order. you just did transpose and put$v^j$in the right side. I do not understand how this can be done, because I think it should be$v^jS_\mathbf{j}^\mathbf{i}=(S^T)_\mathbf{i}^\mathbf{j}v^j.\$ Commented May 15, 2016 at 14:32