Dual of a SDP primal with an extra constraint

Consider the following (standard) primal SDP:

\begin{aligned} \min_{} & \quad \langle{C,X}\rangle \\ {\rm s.t.} & \quad \langle A_i,{X}\rangle = b_i, \\ & \quad X \succeq 0,\\ &\quad Y \succeq 0\ \end{aligned} Except for last PSD constraint, the previous is the standard formulation of a primal SDP problem with decision variable $$X$$. Now, let us consider this extra PSD constraint $$Y \succeq0$$ where $$Y = [Uvec(X)-c]^T\,[Uvec(X)-c]$$ where $$U$$ is some matrix, $$c$$ is a vector and $$vec(X)$$ is the vectorization of $$X$$. Of course this makes $$Y$$ a matrix and we require it is PSD.

Questions:

1. What is the Lagrangian of this problem? For each constraint I am supposed to introduce a "Lagrange multiplier" $$\ell_i$$ which is a dual variable supposedly, so I think I should write something like $$L = \langle{C,X}\rangle + \langle{\ell_1,X}\rangle + \langle{\ell_2,L}\rangle + \sum_i\lambda_i(\langle A_i,X\rangle-b_i)$$ but I cannot figure out if the term $$\langle \ell_2,L\rangle$$ makes sense!
2. How can I write down the dual SDP using the Lagrangian? I assume one defines $$h := \inf \, L$$ and this is equal to $$\lambda^Tb$$ if $$C - \sum_i\lambda_i A_i - \ell_1 -\ell_2 = 0$$. Is this correct? Then by eliminating $$\ell_1+\ell_2$$ one can write down the dual.
3. Can I write down the dual just by looking at the primal without the Lagrangian?

From your definition of $$Y$$, this matrix is necessarily positive semidefinite. So, the constraint on $$Y$$ is superfluous.
1. I am not sure to understand what you mean by $$\langle \ell_2,L\rangle$$ because $$L$$ is your Lagrangian. But in your case, the Lagrangian is pretty standard as you have a standard SDP in primal form. So, we have that
$$L = \langle{C,X}\rangle - \langle{\ell_1,X}\rangle + \sum_i\lambda_i(\langle A_i,X\rangle-b_i).$$
1. The dual SDP is given by $$\max_{\lambda,\ell_1\ge0} \lambda^Tb$$ such that $$\sum_i\lambda A_i-\ell_1=0$$ or, equivalently, $$\sum_i\lambda_i A_i\ge0$$.