I hope you will forgive a math question that comes up in physics contexts where language is loose. This question migrated from Physics SE. I'm finding that I sometimes don't know what kind of product a physics author means. There is a product which can be defined on a vector space that takes two vectors and returns a scalar. There is a product that takes a vector and a covector and returns a scalar.

Is there an agreed-upon language for distinctions between the three words "scalar product", "dot product", and "inner product"?

I can mostly gloss over the language as it's usually clear from context. But not always.

Addition: I think we can all agree what a scalar product is: a map taking two vectors and returning a scalar. A vector space does not have to have a scalar product. To have a scalar product, a vector space needs a metric.

And we can all agree that there is a "contraction" taking one member of a vector space, and one member of it dual, and retuning a scalar, where the dual can be considered to be a scalar-valued linear function on the vector space. Does this "product" have a name?

What I would like to know: what are the definitions of "dot product" and "inner product"?

Addition #2: A commenter (@Hunter) cited a link that points out that physics authors do not all agree on what an inner product is. The following quotation is cited: "If the inner product is taken of two vectors, one must be a contravariant vector and the other a covariant vector. The inner product of two covariant or two contravariant vectors is not defined." [Spherical Astronomy, Robin Michael Green page 495.] Some say the inner product takes $V^*\times V$ into scalars, others say $V \times V$. (Consensus among physicists here is that "inner" = "scalar", i.e. the domain for both is $V \times V$., and "dot" = "scalar", usually reserved for Euclidean geometry.)

Furthermore, most quantum mechanics texts call $\langle\psi\mid\chi\rangle$ an inner product, whereas by my understanding this is a mapping of one bra and one ket to scalars: a contraction of a vector with its dual. I understand that there is an isomorphism in this case, so there is no ambiguity, but the terminology adds to confusion.

One thing is certain: some authors don't tell us which definition they are using, and it's sometimes not clear from context.

Read more: http://www.physicsforums.com/showthread.php?t=735158

  • $\begingroup$ Well that's Hestenes being a little sloppy with his language because I believe it should be called an inner product. In general for the geometric product, a and b can be arbitrary multivectors of different grades. The inner product operation can be shown to be a grade reducing operation. Hestenes himself goes on to show the properties of both operations involved in the geometric product. $\endgroup$ – Elvex Apr 14 '14 at 22:54
  • $\begingroup$ I believe I read in Schutz that the contraction of a vector and co-vector (one-form), $\tilde p(\vec A)$ does not require the intervention of a metric whereas the dot product of two vectors $\vec A \cdot \vec B = \mathbf g(\vec A, \vec B) = \tilde A(\vec B) $ does $\endgroup$ – Alfred Centauri Apr 14 '14 at 23:08
  • $\begingroup$ @AlfredCentauri Isn't that just because one-forms form the dual space, so they map vector to scalars? While the dot product takes two vectors and maps them to a scalar (the metric "transforms" vectors into one-forms) . $\endgroup$ – jinawee Apr 14 '14 at 23:21
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    $\begingroup$ @Hunter Edited. I'll shorten it more if you have suggestions. $\endgroup$ – garyp Apr 15 '14 at 14:00
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    $\begingroup$ @garyp I think you should add in the correct link: physicsforums.com/showthread.php?t=735158, and not: physicsforums.com ;) $\endgroup$ – Hunter Apr 15 '14 at 14:05

Scalar product, inner product and dot product should refer to the same thing, but the last one is normally only used in context of classical vector calculus or if your vector space is $\mathbb R^n$.

Applying a covector to a vector is a special case of tensor contraction that is known as the natural or duality pairing. In the language of differential forms, it is also a special case of the interior product.

The geometric algebra is the Clifford algebra induced by an inner product. As a set, it's the same as the exterior algebra of multivectors, but comes with an additional product - the 'dotless' one.

  • $\begingroup$ Ah! Very clear, but you introduce "interior product". I hesitate to ask, but can you clarify what that is? Also: your third paragraph implies that GA is defined only for metric spaces. Is that correct? And when I see Hestenes use a dot, I should think "scalar product" on a Euclidean space? $\endgroup$ – garyp Apr 14 '14 at 23:58

Let $V$ be a complex vector space. Then the inner product will be a map: \begin{equation} \langle \;\; , \;\; \rangle : V \times V \to \mathbb{C} \end{equation} which is (1) conjugate symmetric in both arguments: \begin{equation} \langle \mathbf{u} , \mathbf{v} \rangle = \overline{\langle \mathbf{v} , \mathbf{u} \rangle} \end{equation} (2) linear in the second argument: \begin{equation} \langle \mathbf{u}, a \mathbf{v} \rangle = a \langle \mathbf{u},\mathbf{v} \rangle \end{equation} where $a$ is some constant, and: \begin{equation} \langle \mathbf{u} , \mathbf{v} + \mathbf{w} \rangle = \langle \mathbf{u} , \mathbf{v} \rangle + \langle \mathbf{u} , \mathbf{w} \rangle \end{equation} (3) the inner product is positive-definite: \begin{equation} \langle \mathbf{u} , \mathbf{u} \rangle \geq 0 \end{equation} and: \begin{equation} \langle \mathbf{u} , \mathbf{u} \rangle = 0 \iff \mathbf{u} = \mathbf{0} \end{equation} It important to note that there are two different conventions of condition (2). The one I used implies that: \begin{equation} \langle \mathbf{u}, a \mathbf{v} \rangle = a \langle \mathbf{u},\mathbf{v} \rangle \implies \langle a \mathbf{u},\mathbf{v} \rangle = \overline{\langle \mathbf{v} , a \mathbf{u} \rangle} = \overline{a \langle \mathbf{v} , \mathbf{u} \rangle} = \overline{a} \langle \mathbf{u},\mathbf{v} \rangle \end{equation}

Finally, note that if the vector spaces are real, then the complex conjugation has no effect and so the inner product is linear in the first and second argument. If the space is $\mathbb{R}^n$, then it becomes the dot product: \begin{equation} \langle \mathbf{u},\mathbf{v}\rangle = \mathbf{u} \cdot \mathbf{v} = \sum\limits_{i=1}^n u^i v^i \end{equation}

Edit: I should also mention that often it is useful to weaken the conditions on the inner product. Namely, to let the inner product not be positive-definite as is the case for the innner product in Minkowski space. I think that this is the inner product that Hestenes is referring to.

  • $\begingroup$ I was with you up until the last two paragraphs (although I would have preferred that you use a different typography for member of the dual space: you use bold face for both spaces). But near the end you say that the inner product (your definition) reduces to the "familiar dot product" (which you did not define). But it's my understanding that the contraction of a vector and a dual vector will never "reduce" to a scalar product. They are two different beasts. A vector space always has a "contraction product" with its dual, but it need not have a scalar product at all. $\endgroup$ – garyp Apr 14 '14 at 23:27
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    $\begingroup$ I thought the definition of inner product was: $\langle \;\; , \;\; \rangle : V \times V \to \mathbb{C} $. No need for dual spaces. $\endgroup$ – jinawee Apr 14 '14 at 23:27
  • $\begingroup$ @jinawee I guess you are right, thanks! I'll fix it now. $\endgroup$ – Hunter Apr 14 '14 at 23:38
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    $\begingroup$ @garyp maybe you will find this link interesting. Either way, nice question because it is something I use a lot intuitively, but I never payed much attention to the exact definition. $\endgroup$ – Hunter Apr 15 '14 at 0:05
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    $\begingroup$ The use of "bilinear" in this answer is incorrect. What the formula says is "linear in the second argument", which together with (1) implies "conjugate linear in the first argument". Together (but without (1)) that is called sesquilinear. Or one could interchange "linear" and "conjugate linear", which is a matter of convention. But under no existing convention this is calld bilinear. $\endgroup$ – Marc van Leeuwen Apr 16 '14 at 11:52

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