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In physics, the definition of a dot (inner) product is often between a vector (“contravariant vector”) and a covector (“covariant vector”). However, in mathematics, a dot product is always defined between two vectors (“contravariant vectors”). What is the mathematical justification for redefining the dot product? I understand why we do it and I know that it works, but I'd like to understand how to justify it from the point of view of mathematics. I'm asking because I've been told that it is not mathematically a priori correct to do it that way.

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  • $\begingroup$ "However, in mathematics, a dot product is always defined between two vectors (“contravariant vectors”)." That really depends on just what field of mathematics you are dealing with. There are lots of formulations that involve a space and its dual. $\endgroup$ – dmckee Feb 1 '14 at 17:17
  • $\begingroup$ "In physics, the definition of a dot (inner) product is often between a vector (“contravariant vector”) and a covector (“covariant vector”)" You do not have to regard things this (confusing) way. Dot product is a function of two vectors in physics too. When you have function of covector and vector, I find it much better to do not call this dot product. This may be called differently, "contraction", "result of action of 1-form on a vector" etc. $\endgroup$ – Ján Lalinský Feb 1 '14 at 17:27
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If you have a finite dimensional real vector space $V$, a (pseudo) scalar product is a bi-linear map $g: V\times V \to \mathbb R$ that is symmetric ($g(u,v)= g(v,u)$) and non-degenerate (if $g(u,v)=0$ for all $v\in V$ then $u=0$).

It is worth remembering for the rest of the discussion that the dual space $V^*$ of $V$ is the space of linear applications $f: V\to \mathbb R$ equipped with the real vector space structure induced by:

$$(af + bg)(v):= af(v)+bf(v)\quad \forall f,g \in V^*, \forall v\in V\:.$$

The elements of $V^*$ are called covariant vectors, while those in $V$ are called contravariant vectors. The real $f(u)$ is also called contraction of $f\in V^*$ and $u\in V$.

The following theorem is the kernel of the two ways to handle the scalar product you mentioned. It proof is elementary and relies upon the given definitions and the fact that the involved vector spaces are finite dimensional (with same dimension).

THEOREM. If $g: V\times V \to \mathbb R$ is a (pseudo) scalar product, then the map $$F: V \ni v \mapsto g(v, \cdot) \in V^*\qquad (1)$$ is a real vector space isomorphism.

$F$ is also known as the musical isomorphism. In view of the given theorem, the scalar product of a pair of contravariant vecotrs, $g(v,u)$, can also be seen as the action of the covariant vector $g(v, \cdot)$ on the contravariant vector $u$:

$$g(v,u) = \left(g(v,\cdot)\right)(u)\:.$$

It is fundamental noticing that the scalar product enters also the RHS. The scalar product of $v$ and $u$ can be seen also as the action of a covariant vector on $u$, but this covariant vector is determined from $v$ just exploiting the existence of the scalar product $g$.

Let us see all this issue in components since this language is more familiar to physicists.

If you fix a basis $e_1,\ldots,e_n$ in $V$, an associated basis $e^{*1}, \ldots, e^{*n}$ arises in $V^*$ called the dual basis. It is completely defined by the requirements: $$e^{i*}(e_j)= \delta^i_j\:.\qquad (2)$$ Using the first basis, if $u=\sum_i v^i e_i$ and $u=\sum_i u^i e_i$, one has: $$g(u,v)= \sum_{i,j} g_{ij}u^iv^j$$ where the matrix of coefficients $g_{ij}$ is symmetric (because $g$ is) and invertible (because $g$ is non-degenerate).

If $f \in V^*$ we can write $f= \sum_i f_i e^{*i}$ using the dual basis of the one exploited in $V$ and, in view of (2) the action of the covariant vector $f$ on the contravariant vector $u$ reads in components:

$$f(v)= \sum_k f_ku^k\:.\qquad (3)$$ Hence (3) is the expression in components of the contraction of $f$ and $v$.

Finally, referring to the given pair of bases, the isomorphism (1) reads: $$F : \sum_i v^ie_i \to \sum_{i,j}v^ig_{ij} e^{*j} =: \sum_j v_j e^{*j}$$ where we have introduced the covariant components of $v$: $$v_k := \sum_i g_{ki}v^k\:.$$ These components are the components of $g(v, \cdot)$.

Putting all together you see the, in components, the scalar product of the contravariant vectors $u$ and $v$ in $V$ can equivalently be written as: $$g(v,u) = \sum_{i,j} g_{ij}v^iu^j = \sum_k v_k u^k$$ where in the last term the covariant components of $v$ show up and the contraction of $g(v, \cdot)$ and $u$ appears written down in components.

So the scalar product cannot be considered as a mere contraction, but is indeed still between vectors of a sort. The musical isomorphism is invoked by the raising or lowering of indices through the metric.

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  • $\begingroup$ This is great. Thank you. I cannot post an answer to my own question for 8 hours, but from what I understand, the answer is that for every vector, there is an associated covector through the musical isomorphism. So the dot product is not merely a contraction, but is indeed still between vectors of a sort. The musical isomorphism is invoked by the raising or lowering of indices through the metric. $\endgroup$ – user125226 Feb 1 '14 at 17:57
  • $\begingroup$ Yes, $u \to g(u,\cdot)$ it is also known as the musical isomorphism. $\endgroup$ – Valter Moretti Feb 1 '14 at 17:59
  • $\begingroup$ I include your comment in my answer. $\endgroup$ – Valter Moretti Feb 1 '14 at 18:05
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The definition of a dot product (that is an operator that takes two objects and produces a scalar) depends on the type of space you're working with. It's just a definition. One does not need any justifications to make a definition except that you hope the definition is useful.

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  • $\begingroup$ This is true, but one hopes that the definition is also compatible with the mathematical theory underlying it. $\endgroup$ – user125226 Feb 1 '14 at 17:53

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