Rotational invariance of Bell state In this lecture on quantum mechanics, Umesh Vazirani shows with a simple calculation that $$\frac{1}{\sqrt{2}}|00\rangle + \frac{1}{\sqrt{2}}|11\rangle = \frac{1}{\sqrt{2}}|uu\rangle + \frac{1}{\sqrt{2}}|u^\perp u^\perp \rangle$$
for any state $u$.  Taking away the physics notation, I translate this as
$$e_1 \otimes e_1 +  e_2 \otimes e_2 = u\otimes u + u^\perp \otimes u^\perp$$
in the the vector space $V \otimes V$, $V = \mathbb{C}^2$, where $e_1,e_2$ is the standard basis of $\mathbb{C}^2$ and $u, u^\perp$ is any another basis attained by applying a (real) rotation to $e_1, e_2$.  Is there a coordinate-free or more conceptual proof than the one Vazirani gives?  What's the geometric meaning of this vector? Can we generalize?
 A: Remark 1: In quantum mechanics, the most general transformation mapping pure states of a two-level quantum system (aka qubit) to other pure states is represented as a unitary matrix in $SU(2)$. However, as Vazirani points out in the video and you note in your question, his proof only applies to real rotations, i.e. matrices in $SO(2) \subset SU(2)$.

Proof: Let $\beta_{00} = |00\rangle + |11\rangle$ where we ignore normalization since vector norm is preserved by unitary transformations. A key property of $\beta_{00}$ which we need is that for any $A \in GL(2, \mathbb{C})$
$$
(A \otimes I) \beta_{00} = (I \otimes A^T) \beta_{00}\tag1.
$$
This can be shown easily by writing it out in the matrix notation or by following indices in the tensor notation. However, since you requested a coordinate-free or more conceptual proof, we will instead use tensor networks employing some basic properties shown in this paper. Note that the tensor network representation of $\beta_{00}$ is a simple bend (so called "cup") and $(1)$ is merely a restatement of equation $(46)$ in the paper.
Now, let $U \in SO(2)$ be the real rotation that maps $|0\rangle$ to $|u\rangle$ and $|1\rangle$ to $|u^\perp\rangle$. Applying it to both qubits and using $(1)$ we obtain
$$
(U \otimes U) \beta_{00} = (I \otimes U^TU) \beta_{00}
$$
but $U$ is orthogonal so
$$
(U \otimes U) \beta_{00} = \beta_{00}.
$$
On the other hand,
$$
(U \otimes U) \beta_{00} = |uu\rangle + |u^\perp u^\perp\rangle.
$$
Comparing the last two equations, we obtain the desired rotational invariance.

Remark 2: For a generic unitary $U$ the proof above fails, because $U^TU$ is not necessarily identity. For a concrete counterexample consider $U=S$ where
$$
S = \begin{pmatrix}
1 & 0\\
0 & i
\end{pmatrix}
$$
where we obtain
$$
(S \otimes S)\beta_{00} = |00\rangle - |11\rangle \ne \beta_{00}.
$$

Remark 3: We can generalize the above result if we replace $\beta_{00}$ with a different Bell state. In his video Vazirani mentions that $\beta_{11} = |01\rangle - |10\rangle$ is invariant under general unitary transformations. Let us see how this happens.
First, note that $\beta_{11} = (Z \otimes X)\beta_{00}$ where
$$
Z = \begin{pmatrix}
1 & 0 \\
0 & -1
\end{pmatrix} \,\,\,\,
X = \begin{pmatrix}
0 & 1 \\
1 & 0
\end{pmatrix}.
$$
From equation $(45)$ in the paper linked above, we see that $\beta_{11}$ is the Levi-Civita tensor in two dimensions. An important property of this tensor, expressed in $(12)$ in the paper, is
$$
(U \otimes U)\beta_{11} = (\det U) \beta_{11}.
$$
Thus, for $U \in SU(2)$ we have
$$
(U \otimes U)\beta_{11} = \beta_{11}
$$
so $\beta_{11}$ is invariant under arbitrary unitary transformations in $SU(2)$ applied to both qubits.
