# Inertia tensor in canonical and rotated basis

So say we can describe some angular velocity $$\boldsymbol{\omega}$$ in canonical basis, $$\{\mathbf{e}_i\}$$, and a rotated basis, $$\{\mathbf{\tilde{e}}_i\}$$, like

$$\boldsymbol{\omega}=\omega_1\mathbf{e}_1+\omega_2\mathbf{e}_2+\omega_3\mathbf{e}_3=\tilde{\omega}_1\mathbf{\tilde{e}}_1+\tilde{\omega}_2\mathbf{\tilde{e}}_2+\tilde{\omega}_3\mathbf{\tilde{e}}_3,$$

and its coordinates in each basis and their basis are related by the rotational matrix $$\mathbf{R}$$ in the following way: \begin{aligned} \begin{pmatrix} \tilde{\omega}_1\\ \tilde{\omega}_1\\ \tilde{\omega}_1 \end{pmatrix}=\mathbf{R} \begin{pmatrix} \omega_1\\ \omega_2\\ \omega_3 \end{pmatrix} \end{aligned}\ \text{and}\ \begin{aligned} \begin{pmatrix} \mathbf{\tilde{e}}_1\\ \mathbf{\tilde{e}}_2\\ \mathbf{\tilde{e}}_3 \end{pmatrix}=\mathbf{(R^{-1})} \begin{pmatrix} \mathbf{e}_1\\ \mathbf{e}_2\\ \mathbf{e}_3 \end{pmatrix}. \end{aligned}

Then, what I want to ask is whether when calculating the angular momentum in canonical basis, $$\mathbf{L_e}=\mathcal{I_e}\boldsymbol{\omega_{e}}$$ and in rotated basis $$\mathbf{L_{\tilde{e}}}=\mathcal{I_{\tilde{e}}}\boldsymbol{\omega_{\tilde{e}}},$$

would $$\mathcal{I_e}=\mathcal{I_{\tilde{e}}}$$?

Because I thought that maybe if $$(\mathcal{I_{ij}})_e=\iiint_V \rho(r^2\delta_{ij}-x_ix_j) dV$$

then $$(\mathcal{I_{ij}})_{\tilde{e}}=\iiint_\tilde{V} \rho(r^2\delta_{ij}-\tilde{x}_i\tilde{x}_j) d\tilde{V}.$$

Note that $$\mathbf{\tilde{x}}=\mathbf{Rx}.$$

$$(I_{\tilde e})_{ij} = \int_V \rho (\tilde r^2 \delta_{ij}-\tilde x_i \tilde x _j)$$ of course $$r^2 = \tilde{\mathbf{r}} \cdot \tilde{\mathbf{r} } = (\mathbf{R}\mathbf{r})^{\mathrm{tr}}\cdot \mathbf{R} \mathbf{r} = \mathbf{r}\mathbf{R}^{-1}\mathbf{R}\mathbf{r}=\mathbf{r}\cdot\mathbf{r}$$. This is true for any scalar, also the square of the angular momentum. Let us go back to the formula $$(I_{\tilde e})_{ij} = \int_V \rho (\tilde r^2 \delta_{ij}-R_{ia}x_a\,R_{jb} x_b)$$ which differs from $$(I_{e})_{ij}$$.