Fundamental form on a vector space  
Apologies for posting picture instead of typing as there is some problem with keyboard. 
I am confused with notation $w\in \Lambda^2 V^*$. Does it mean we consider $\omega$ as $:V\times V\rightarrow \mathbb{R}$  in which case it should be an element of $\Lambda^2V$. so, I guess they  consider extension of $\omega$ to $V^*$ i.e., map $V^*\times V^*\rightarrow \mathbb{R}$ but then did not specify what extension it is. I defined $\omega(f,g)=\left< u,v\right>$ where $u$ is such that $f(x)=\left<x,u\right>$ for all $x\in V$ and $v\in V$ is such that $g(x)=\left<x,v\right>$ for all $x\in V$. But proof says

since $$\omega(w,v)=\left<Iw,v\right>=-\left<w,Iv\right>=-\left<Iv,w\right>=-\omega(v,w)$$

This only says $\omega\in \Lambda^2 V$. Defining $\omega\left<f,g\right>$ as i have defined, I proved that $\omega\in \Lambda^2 V^*$. I am confused with the notation. It would be grateful if any one who has used this book can clarify this.
 A: The book is correct. 
A skew-symmetric bilinear map $V\times V \to \mathbb{R}$ descends to a linear map $\bigwedge^2V \to \mathbb{R}$, i.e. an element of $\left(\bigwedge^2V\right)^* \cong \bigwedge^2V^*$.

As for your question in the comments (why is $\omega \in \bigwedge^{1,1}V^*$?), recall that ${\bf I} : \bigwedge^*V_{\mathbb{C}} \to \bigwedge^*V_{\mathbb{C}}$ is the linear operator that acts on $\bigwedge^{p,q}V$ by multiplication with $i^{p-q}$, i.e. 
$${\bf I} = \displaystyle\sum_{p,q}i^{p-q}\,\Pi^{p,q}$$
where $\Pi^{p,q} : \bigwedge^*V_{\mathbb{C}} \to \bigwedge^{p,q}V$ is the natural projection.
There is a corresponding operator on the dual space $\bigwedge^*V_{\mathbb{C}}$, also denoted ${\bf I}$, given by ${\bf I}(\alpha)(v_1, \dots, v_k) = \alpha({\bf I}(v_1), \dots, {\bf I}(v_k))$. If $\alpha$ is a $(p, q)$-form, then ${\bf I}(\alpha) = i^{p-q}\alpha$ so again we have
$${\bf I} = \displaystyle\sum_{p,q}i^{p-q}\,\Pi^{p,q}$$
where $\Pi^{p,q}$ now denotes the natural projection $\bigwedge^*V_{\mathbb{C}}^* \to \bigwedge^{p,q}V^*$.
For $\alpha \in \bigwedge^2V_{\mathbb{C}}^*$, we have $\alpha = \alpha^{2,0} + \alpha^{1,1} + \alpha^{0,2}$ where the exponent denotes the bidegree. So
$${\bf I}(\alpha) = i^{2-0}\alpha^{2,0} + i^{1-1}\alpha^{1,1} + i^{0-2}\alpha^{0,2} = -\alpha^{2,0} + \alpha^{1,1} - \alpha^{2,0}.$$
If ${\bf I}(\alpha) = \alpha$, then by comparing bidegrees, we see that $\alpha^{2,0} = 0$ and $\alpha^{0,2} = 0$, so $\alpha = \alpha^{1,1} \in \bigwedge^{1,1}V^*$. In particular, $\omega \in \bigwedge^{1,1}V^*$.
