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If $M$ is $2n+1$ dimensional manifold, and $M'= M\times \mathbb R$ Let $x_1,y_1,... x_n, y_n,t', t$ be coordiante of $M'$. With $t$ for coordinate for $\mathbb R$. Let $$ \omega= \sum_{i=1}^n dx_i\wedge dy_i+ dt'\wedge dt$$ $i: M\to M'$ be inclusion. How to calculate $i^*(\omega )$.

By definition: for $v, w\in T_p M$, we should have $$\alpha_p(v,w)= i^*\omega(v,w)= \omega_{i(p)}(di_p(v), di_p(w))$$ $$\alpha_p(v,w)= \left(\sum_{i=1}^n dx_i\wedge dy_i+ dt'\wedge dt\right)\left(\sum_{j=1}^nv_j\frac{\partial}{\partial x_j}+v_0\frac{\partial}{\partial t'},\sum_{j=1}^nw_j\frac{\partial}{\partial x_j}+w_0\frac{\partial}{\partial t'}\right) $$ After that i am getting confuse... How to write $\alpha$. I want an expression for $\alpha$.

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You can use linearity of the pullback, and its compatibility with $d$. It remains to compute $i^*(dt)$. – Davide Giraudo May 30 '12 at 9:34
$i^*(dt)=0$ and $i^*(dx_i)= dx_i$, $i^*(dy_i)= dy_i$, $i^*(dt')= dt'$. so whether we have: $i^*(\omega)= \sum_{i=1}^n dx_i\wedge dy_i$ Is this logic reasonable. – Junu May 30 '12 at 9:41
Yes. You can answer your own question, then it won't be unanswered anymore. – Davide Giraudo Jun 16 '12 at 20:44

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