When we take a dot product between two vectors of a vector space, we actually "act" by a 1-form (dual vector) on a vector. So why most books define the dot product between vectors? Of course with the help of a metric we can take a dot product between two vectors, but technically the metric converts one of the two vectors to a 1-form.

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    $\begingroup$ What?${}{}{}{}$ $\endgroup$ – Git Gud Mar 23 '13 at 19:40
  • $\begingroup$ $\bf{v}\cdot \bf{u}$ $=g_{ij}v^i u^j=v_j u^j$. Where $v_j$ are the components of the covector (1-form) v and $u^j$ of the vector u. $\endgroup$ – AndyK Mar 23 '13 at 19:49
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    $\begingroup$ I hate to break you the news, but tensor analysis on manifolds is not the only branch of mathematics, where vectors are used. Elsewhere (may be 95%+ of applications of vector spaces?) it makes sense to define an inner product of two vectors, but the concept of a 1-form is largely meaningless. $\endgroup$ – Jyrki Lahtonen Mar 23 '13 at 19:58
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    $\begingroup$ I'm pretty sure that analysts are not willing to discard the notion of inner product since they love $L^p$ spaces so much. Indeed, they exploit the inner product notation even when $V^*$ is not isomorphic to $V$! $\endgroup$ – Sangchul Lee Mar 23 '13 at 20:08

An inner product is definitely a certain bilinear map $$\langle \cdot, \cdot \rangle: V \times V \longrightarrow \Bbb R$$ which takes two vectors as its arguments. If you're thinking in terms of a metric tensor $g$, then $g$ is a type $(0,2)$-tensor, which means it has two vector arguments (no covector arguments).

What you're thinking of is the fact that a choice of inner product $\langle \cdot, \cdot \rangle$ on a vector space $V$ gives an isomorphism $$\varphi: V \longrightarrow V^\ast,$$ $$v \mapsto \langle v, \cdot \rangle.$$ In this sense we can identify the inner product as the action of a covector on a vector: $$\langle u, v \rangle = \varphi(u)(v).$$ In terms of components of the metric tensor, this is equivalent to $$\langle u, v \rangle = g_{ij} u^i v^j = u_j v^j.$$ An element of $V^\ast$ acts on $V$ in the same way no matter what basis we choose, but there is no canonical identification of $V$ with $V^\ast$. Without a metric, the lowered index components $u_i$ of a vector $u$ make no sense.

  • $\begingroup$ I know that we define the inner product as a bilinear map from $V \times V$ to $\mathbb{R}$. But we can also define the inner product of $u,v\in V$ by taking the dual vector of $u$ (or $v$), i.e. the linear functional on $V$ which "sends" $u$ to $||u||^2\in \mathbb{R}$, and then acting with it on the vector $v$. Isn't the second definition more fundamental? $\endgroup$ – AndyK Mar 23 '13 at 20:28
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    $\begingroup$ @Andyk: What do you mean by $\|u\|^2$ when you have no special choice of inner product or norm? Is it the standard $u_1^2 + \cdots + u_n^2$, where $u_i$ are the coordinates of $u$ with respect to some chosen basis for $V$? Even if this is the case, there is not a unique linear functional which sends some fixed vector $u$ to $\|u\|^2$ (or whatever fixed value you want). For example, the linear functionals $(a,b) \mapsto a$ and $(a,b) \mapsto a+b$ on $\Bbb R^2$ both send $(1,0)$ to $\|(1,0)\| = 1$, but they are not the same functional. $\endgroup$ – Henry T. Horton Mar 23 '13 at 21:26

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