Natural way to extend an inner product on a vector space to its dual Let $V$ be a finite dimensional vector space over $\mathbb{R}$ and $\left<~,\right>$ be a scalar product on $V$. 
Is there any obvious extension of $\left<~,\right>$ to $V^*$ the dual of $V$.
Is there any obvious extension of $\left<~,\right>$ to $\Lambda^k V$ the $k$ th exterior power of $V$.
I was thinking about the assignment $\left<f,g\right>=\left<u,v\right>$ where 
$u\in V$ such that $f(x)=\left<x,u\right>$ for all $x\in V$ and $g(x)=\left<x,v\right>$ for all $x\in V$ which exists by Riesz Representation theorem. I am not quite sure if this is what the author is referring to.
For $w_1\wedge w_2\cdots\wedge w_k,v_1\wedge v_2\cdots\wedge v_k$  define 
$\left<w_1\wedge w_2\cdots\wedge w_k,v_1\wedge v_2\cdots\wedge v_k\right>=\sum_{i,j}(-1)^{i+j}\left<w_i,v_j\right>$. I am not quite sure if this is what the author is referring to.
This came when I was reading Complex Geometry - An Introduction by Daniel Huybrechts.
Any suggestion is welcome
 A: Your suggestion for $V^{*}$ is indeed a natural one and often used. A metric on $V$ gives you an isomorphism $T\colon V \rightarrow V^{*}$ (using Riesz Representation Theorem) and then you put an inner product on $V^{*}$ which turns $\phi$ into an isometry (so you define $\left< \varphi, \psi\right>_{V^{*}} := \left< T^{-1}(\varphi), T^{-1}(\psi) \right>_V$). Note that this means that if $(e_1,\dots,e_n)$ is an orthonormal basis then $(e^1 = T(e_1), \dots, e^n = T(e_n))$ is the dual basis of $(e_1,\dots,e_n)$ and so we see that the inner product on $V^{*}$ makes $e^1,\dots,e^n$ into an orthonormal basis of $V^{*}$.
Regarding the exterior powers $\Lambda^k(V)$, a natural choice is to define an inner product on decomposable $k$-wedges by the formula
$$ \left< v_1 \wedge \dots \wedge v_k, w_1 \wedge \dots \wedge w_k \right>_{\Lambda^k(V)} := \det \left( \left< v_i, w_j \right> \right)_{i,j=1}^k $$
and then to extend using bilinearity to all $\Lambda^k(V)$. The reason this is natural is that if $(e_1,\dots,e_n)$ is an orthonormal basis for $V$, then $(e_{i_1} \wedge \dots \wedge e_{i_k})_{I = (i_1 < \dots < i_k)}$ is an orthonormal basis for $\Lambda^k(V)$.
