Is $\mathbb{C}[T \times_T T] = (\mathbb{C}[T] \otimes \mathbb{C}[T])^T$? Let $T$ be an algebraic group. There is a left $T$ action on $T$ given by left multiplication and a right action on $T$ given by right multiplication. Let $T$ acts in the middle of $T \times T$ by the formula: $t(t_1, t_2) = (t_1 t^{-1}, t t_2)$. Then there is a corresponding middle action of $T$ on $\mathbb{C}[T \times T] = \mathbb{C}[T] \otimes \mathbb{C}[T]$. Let $(\mathbb{C}[T] \otimes \mathbb{C}[T])^T = (\mathbb{C}[T \times T])^T$ be the set of invariants of the middle $T$ action. Is $\mathbb{C}[T \times_T T] \cong (\mathbb{C}[T] \otimes \mathbb{C}[T])^T$? Here $T \times_T T$ is the fiber product with two appropriate maps from $T$ to $T$. Thank you very much.
Edit: the two appropriate maps from $T$ to $T$ are both identity map: $t \mapsto t$.
 A: If $G$ is a abstract group, there is a map $\beta:(x,y)\in G\times G\mapsto (xy,y),G\times G$ which is a bijection, with inverse $(x,y)\mapsto(xy^{-1},y)$. If we let $G$ act in the middle on the domain of $\beta$, as in your question, so that $t\cdot(x,y)=(xt^{-1},ty)$, and on the codomain of $\beta$ so that $t\cdot(x,y)=(x,ty)$, then the map $\beta$ is a $G$-equivariant bijection.
It follows from this that the quotient $(G\times G)/G$ with the action as in the domain of $\beta$ is in bijction with the quotient $(G\times G)/G$ with action as in the codomain of $\beta$. The latter is clearly the sam as $G\times(G/G)$ with $G/G$ denoting the quotient of $G$ by the translation action of $G$ on the left. Of course G/G$ is just a point.
If $G$ is an affine algebraic group, then all of this is true in the algebraic sense. This means that there is an isomorphism $(G\times G)/G$ with action in the middle, with $G$, as affine varieties.
Your question is the dual perspective of this, in terms of rings of regular functions.
