The Liouville form pulled back to the Lagrangian on the cotangent bundle vanishes

Question

Consider the cotangent bundle of a smooth manifold, with the zero section removed, $\dot{T}^*M$. Let $i : L\hookrightarrow \dot{T}^*M$ be a conic Lagrangian submanifold(i.e. $(x, \xi)\in L\implies (x,t\xi)\in L$ for $t>0$). Denote $$\lambda=\sum_{i=1}^n\xi_idx_i$$ be the Liouville $1$-form such that $d\lambda$ is symplectic. The claim is $i^*\lambda=0$. It's unclear to me why this should even be true. Because $L$ is Lagrangian, $i^*(d\lambda)=0$. But I am not sure why the one form even pulls back to zero. In the case $L=graph(df)$ I can't even do it. I can show it pulls back $d\lambda$ to zero as once you differentiate we get sums of $\frac{\partial^2f}{\partial x_i\partial x_j}dx_i\wedge dx_j+\frac{\partial^2f}{\partial x_j\partial x_i}dx_j\wedge dx_i$, which are all zero but I don't know how to show it for $\lambda$. Any help for this problem?

Answer 1

The claim simply isn't true. The defining property of the Liouville form is that $\alpha^*\lambda=\alpha$ for any one form $\alpha$ on $M$ (viewed as a map $M\to T^*M$). This means that the graph of any closed one form is Lagrangian, as $\alpha^*(d\lambda) =d\alpha^*\lambda=d\alpha=0.$

For a non-zero closed one form we have $\alpha^*\lambda=\alpha\neq 0$