Cfrgtkky

I would like to see the fact that the components of a vector transform differently (controvariant transformation) than the unit bases vectors (covariant transformation) for the specific case of cartesian to polar coordinate transformation.

The polar unit vectors $hat{r}$ and $hat{theta}$ can be expressed in terms of cartesian unit vectors, $hat{x}$ and $hat{y}$, as the following
begin{equation}
hat{r}= text{cos}phi hat{x} + text{sin}phi hat{y} \
hat{theta}= -text{sin}phi hat{x} + text{cos}phi hat{y} tag{1}
end{equation}

Any vector, $vec{V}$, can be expressed in the cartesian coordinate system as $vec{V}=V_x hat{x} + V_y hat{y}$. The same vector can be expressed in polar coordinates as $vec{V}=V_r hat{r} + V_theta hat{theta}$. We then have
begin{equation}
V_x hat{x} + V_y hat{y}=V_r hat{r} + V_theta hat{theta}. tag{2}
end{equation}
I then project both sides of (2) once onto $hat{r}$, and once onto $hat{theta}$. Using (1) and (2) we get
begin{equation}
V_r= text{cos}phi V_x+text{sin}phi V_y \
V_theta= -text{sin}phi V_x+text{cos}phi V_y tag{3}
end{equation}

Comparing (1) and (3), both the unit vectors and the components of a vector are transforming with the same rule, which is a contradiction! What am I missing here?

As @TedShifrin pointed out, the correct transformation for the dual is $(A^{-1})^T$.

For instance, in special relativity contravariant vectors transform as

$ V^{mu '} = Lambda ^{mu '} _{,nu} V^{nu}$,

where $ Lambda ^{mu '} _{,nu} $ is the Lorentz transformation taking component from the unprimed frame to the primed frame. For covariant components we have to use the inverse of the Lorentz transformation:

$ V_{mu '} = big( Lambda^{-1} big)^{nu } _{,mu '} V_{nu}$.

If we assume that $V'$ and $V$ are column vectors representing, $ V_{mu '}$ and $V_{nu}$, respectively. Then in matrix format we have:

$ V' = big( Lambda^{-1} big)^T V$.

This can also be found here: https://en.wikipedia.org/wiki/Lorentz_transformation

Of course, one can get to the same conclusion in the context of group theory. If the vector $x$ transforms according to $x rightarrow x'=Ax$ where $A$ is the group member. Then the dual $tilde{x}$ transforms as $tilde{x} rightarrow tilde{x} '= big(A^{-1} big)^Ttilde{x}$ so that $tilde{x}^T x$ is an invariant in all frames.