I'm very confused about the tensor representation of Control gates. I had thought that all control gates were of the form $P_0 \otimes I+P_1\otimes \hat{A}$, where this does nothing to the first qubit and acts with the operator $A$ on the second, but that there was a conditional implication involved. And I thought that $I\otimes A$ just meant that you act on the second and not the first but there was no implication of of conditionality?
Could someone please clear this up for me?
The idea behind a controlled gate is that nothing (i.e. $I$) is performed on the control qubit, but the action on the target qubit depends on the state of the control qubit.
Therefore, you can't write the action on the target qubit as a unitary operation on only that qubit. That is to say, the controlled operation can not be of the form $I_{c} \otimes A_{t}$.
However, the different specific actions on the target qubit can be 'entangled' to the different states of the control qubit. This is exactly where your notation comes in. Consider the $CNOT$ operation: if the control qubit's state is $|0\rangle$, we perform $I$ on the target qubit, and if the state is $|1\rangle$, we perform $X$. This becomes: \begin{equation} CNOT = |0\rangle\langle0|\otimes I + |1\rangle\langle1|\otimes X, \end{equation}
so in this case $P_{0} = |0\rangle\langle0|$ and $P_{1} = |1\rangle\langle1|$.
Note that we always have $\sum_{i} P_{i} = I$ if we sum over a whole basis. Therefore, the action on the control qubit, without any regard to the target qubit, is indeed always $I$, i.e. nothing.
Edit: as an added note, keep in mind that these projector operators $P_{i}$ can never correspond to unitary operations.
