# Why do basis vectors transform covariantly?

I have some trouble understanding transformation rules of basis vectors. My question/goal is to obtain a mathematical derivation to see why basis vectors transform covariantly and other vectors (components) contravariantly. I have three questions.

• Question 1: why can't I use the vector component transformation equation (with $$R_{ji}$$) to transform each components of the old basis vectors into the new ones? It should work for any vector right?

• Question 2: why is $$R_{ji}$$ sometimes expressed as $$\frac{\partial x^{'i}}{\partial x^j}$$ as I have read in some sources?

• Question 3: as the change of basis should be independent of any vector in particular why do I still have vector component terms in my derivation?

As I have learned the components of a vector transform under coordinate transformation as

$$\begin{equation} V'^i=\sum_j R_{ji}V^j \end{equation}$$

where $$R_{ij}$$ is a rotation matrix.

Now of course the vector V itself is a geometrical object and independent of coordinates. The vector should be defined as follows:

$$\begin{equation} \boldsymbol{\vec{V}}=V^i\boldsymbol{\hat{e}_i} \end{equation}$$

So I tried to express the vector in both old coordinate terms and new ones:

$$\begin{equation} \boldsymbol{\vec{V}}=V^i\boldsymbol{\hat{e}_i}=V'^i\boldsymbol{\hat{e}^{'}_i} \end{equation}$$

$$\begin{equation} \sum_j R_{ji}V^j \boldsymbol{\hat{e}^{'}_i}=V^i\boldsymbol{\hat{e}_i} \end{equation}$$

And therefore

$$\begin{equation} \boldsymbol{\hat{e}^{'}_i}=\frac{V^i}{\sum_j R_{ji}V^j }\boldsymbol{\hat{e}_i} \end{equation}$$

This seems strange since the new basis vector should not depend on any particular vector. When I try to set the components of the vector equal to one in order to just leave us with the basis vectors I find the following expression:

$$\begin{equation} \boldsymbol{\hat{e}^{'}_i}=\frac{1}{\sum_j R_{ji}}\boldsymbol{\hat{e}_i} \end{equation}$$

is this equal to $$R_{ij}$$ and would this imply covariant transformation? I have a big feeling that my derivation is very wrong since my knowledge lacks at linear algebra.

I hope someone can answer (some of) these questions. Thank you in advance!

-Jesse

• "Covariant" just means "varies like the basis vectors" and contravariant means "varies opposite to the basis vectors". It's no wonder basis vectors are covariant, basis vectors vary like the basis vectors. Mar 24, 2021 at 21:42
• @ChemiCalChems okay thank you, then I guess my question is why to basis vectors transform the way they do? How do basis vectors vary? Mar 24, 2021 at 21:44
• I think your conclusion after "and therefore" is wrong (there is a sum over "i"). Mar 24, 2021 at 21:52
• @heaven-of-intensity right that's also what I suspected. However, since the vectors are fundamentally equal, the first component should be the same when multiplying everything out. So when looking only at i=1 it should work and should evade the summation problem. The problem is that there still remain vector components. Is this a valid method? Mar 24, 2021 at 22:03
• @jessegerritsen When you set i=1 on the RHS, you cannot set i=1 on the LHS! The "i"'s are not the same variables, so you cannot "divide" out. For example, assume $e_1+e_2=f e_1+ g e_2$. Mar 24, 2021 at 22:07

First you need to decide whether you want to use sum symbols or whether you want Einstein summation convention (implicit sums).

The vector decomposition you have written down is actually Einstein: $$\vec V = V^i \hat e_i := \sum_i V^i \hat e_i$$

Second, you should keep upper indices upper, and lower indices lower, on both sides. Rather than writing (wrongly, so to say) $$V^{\prime i}=\sum_j R_{ji}V^j$$ you should write $$V^{\prime i}=\sum_j R^i_jV^j$$ because then $$i$$ is upper on both sides. Upper or lower positions indicates its covariance/contravariance which makes checking correctness/consistency of your calculations much easier.

Third, in the equation $$\sum_j R_{ij} V^j \hat e^{\prime}_i = V^i \hat e_i$$ you have combined the former mistakes in the most unfortunate ways. The correct (verbose) expression is $$\sum_{i}\sum_{j} R^i_j V^j \hat e^{\prime}_i = \sum_{i} V^i \hat e_i$$ That is why you can't just divide by the sum over $$j$$ an rearrange them to the other side. But what you actually can derive of this equation is the transformation behavior of the basis vectors. If you compare the left hand side and the right hand side (maybe by swapping/renaming indices $$i,j$$ on the left hand side, to make it clearer), you see that $$\sum_j R^j_i \hat e^{\prime}_j = \hat e_i$$ must be satisfied in order for the former equation to hold true. Note, however, that in the other transformation direction you have to use the inverse transform $$\hat e^{\prime}_j = \sum_i (R^{-1})^i_j \hat e_i$$ where $$R^{-1}$$ is defined such that $$R^{-1}R=I$$ (I=unity matrix), or $$\sum_j (R^{-1})^i_j R^j_k=\delta^i_k$$ Only for orthogonal transforms is the inverse of $$R$$ equal to its transpose.

As to question number 2, this is usually only used when the transformation is nonlinear (although you may use it for linear transforms as well). Since $$x^{\prime i} = x^{\prime i}(x)$$ the derivative of any function $$f(x^{\prime})$$ can be expressed by the chain rule of differentation: $$\frac{\partial f}{\partial x^j}=\sum_i \frac{\partial f}{\partial x^{\prime i}}\frac{\partial x^{\prime i}}{\partial x^j}=\sum_i \frac{\partial f}{\partial x^{\prime i}}R^i_j$$

• Thank you for your answer! This is makes it much clearer. Mar 24, 2021 at 22:13
• You're welcome. Mar 24, 2021 at 22:13
• so the only question that I still have is how should I transform a basis vector given a rotation matrix or set of derivatives. Could you maybe explain that? Mar 24, 2021 at 22:15
• @jessegerritsen: See my edit after boldfaced "can". Mar 24, 2021 at 22:21
• yes this is exactly what I was looking for, again thanks! Mar 24, 2021 at 22:26