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Consider $\mathbb{R}^n$ as a differentiable manifold with the standard coordinates $x^j: \mathbb{R}^{n} \to \mathbb{R}, x_1 e^1 + ... + p_n e^n \mapsto p_j.$

The vector field $X_A$, which in these coordinates takes the form $$X_A(p) = A_{ij} x^j(p) \frac{\partial}{\partial x^i}_p$$ corresponds simply to the ODE $$\frac{d}{dt} x = Ax.$$

Now I want to understand what happens to the vector field $X_A$ under a change of coordinates of the form $$y = Cx.$$ In my intuition, this should correspond to a change of basis in the sense of linear algebra, that is I would expect in these new coordinates the vector field is of the form $X_B$, with $B = CAC^{-1}$.

However, when I do the calculations, I obtain a different results:

We have $$\frac{\partial}{\partial x^i} = \frac{\partial x^i}{\partial y^k}\frac{\partial}{\partial y^k},$$ hence since $x=C^{-1}y$, $$\frac{\partial}{\partial x^i} = C^{-1}_{ik} \frac{\partial}{\partial y^k}.$$

Plugging this into the definition of $X_A$, one obtains

$$X_A(p) = A_{ij} C^{-1}_{jl} y^l C^{-1}_{ik} \frac{\partial}{\partial y^k}= ({C^{-1}}^T AC^{-1})_{kl} y^l \frac{\partial}{\partial y^k} = X_{C^{-T}AC^{-1}}$$

This seems to suggest that the transformation behaviour is not as I thought above.

Now, am I making a mistake in the calculations here, or does a linear ODE under coordinate change not behave in the same way as a change of basis? If the latter, can someone give an intuitive explanation of the difference? Also, is there some transformation of ODE's that does behave in the way I expected?

Edit:

After reading the comments, I realize I might have been implicitly using some wrong assumption of independence of coordinates and basis, when simply replacing $x$ by $C^{-1}$ in the equation for the vector field instead of trying to solve for the coordinates in the new basis.

Thus, we posit $$X_A = A_{ij} x^j \frac{\partial}{\partial x^i} = d^k \frac{\partial}{\partial y^k}=d^k \frac{\partial y^k}{\partial x^i} \frac{\partial}{\partial x^i}=d^k C_{ki} {\partial x^i},$$ or, in matrix-vector notation

$$Ax \frac{\partial}{\partial x} = C^T d \frac{\partial}{\partial x}.$$

This should lead to $d = C^{-T}Ax$. But now we should be able to plug in $x = C^{-1}y$ here. Then the equation still reads

$$X_A = C^{-T}AC^{-1}y \frac{\partial}{\partial y}.$$

a_student
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  • I believe you fell into the covariant/contravariant trap by which components of vectors transform with the inverse of the matrix that we use to transform basis vectors. Apart from this everything seems OK. – Kurt G. Sep 23 '23 at 11:37
  • @KurtG. can you explain at which point I fell in that trap? Is the formula for expressing $\frac{\partial}{\partial x^i}$ wrong? – a_student Sep 23 '23 at 15:50
  • Since you are using the notation fluently I think for now a bigger picture is what you probably just need: all coordinate transformations can be derived from the chain rule. Plus from the requirement that a vector $$ \boldsymbol{V}=V^i\partial_{x_i}=U^i\partial_{y_i} $$ must not change under coordinate transformations you can figure out how its components $V^i$ transform into $U^i,.$ – Kurt G. Sep 23 '23 at 15:59
  • You will find that they are contravariant when you baptize the transformation of the basis vectors $\partial_{x_i}$ to $\partial_{y_i}$ covariant. In matrix language: you switch to the inverse transformation matrix. Lastly: I think in linear algebra we transform components contravariantly. – Kurt G. Sep 23 '23 at 15:59
  • @KurtG. I tried your suggestion but I still get the same result, see the edit above – a_student Sep 24 '23 at 20:06
  • Your equation $$\frac{\partial}{\partial x^i} = \frac{\partial x^i}{\partial y^k}\frac{\partial}{\partial y^k}$$ looks wrong. Numerator and denominator should be flipped. $y^k$ must "cancel". – Kurt G. Sep 25 '23 at 08:54
  • @KurtG. That was the error! I think I used the same mnemonic about canceling, but at some point I used the notation $\partial x^i$ for $\frac{\partial}{\partial x^i}$, and then the same trick leads to the wrong formula! If you want to post an answer, I can accept it. – a_student Sep 25 '23 at 09:21
  • No thanks. Too much work. Glad that I could help. – Kurt G. Sep 25 '23 at 09:40

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