$$a\cdot b$$


$$\langle a,b\rangle$$

$$\langle a\mid b\rangle$$





Matrices are a different type of object from vectors, so the transpose operation deserves a separate symbol $^T$.

An abstract vector space may have several different inner products, which must have distinct names, so $g$ is reasonable.

But all the others seem superfluous compared to the dot. The notation $a\cdot b$ correctly suggests bilinearity, that it acts like multiplication. It can also be used in an abstract (no inner product) space for the action of a covector: $\alpha(b)=\alpha\cdot b$

  • $\begingroup$ The notation $\langle a | b \rangle$ is quite practical in QM, since it allows you to really emphasize the duality between bras $\langle a | $ and kets $| b \rangle$. Other than that, whenever I see $a \cdot b$ I think of the Euclidean scalar product. I generally use $\langle a, b \rangle$ when working with general inner products. $\endgroup$ – MisterRiemann Sep 10 '18 at 18:38
  • $\begingroup$ On the other hand $a\cdot b$ can denote so many different algebra products, e.g., see here. Of course we can use $ab$, $a\circ b$, or $a\bullet b$ etc. You see, the problem can also be the other way around. $\endgroup$ – Dietrich Burde Sep 10 '18 at 18:42
  • $\begingroup$ So many different names? Surely because it arises in so many different contexts. $\endgroup$ – Lord Shark the Unknown Sep 10 '18 at 18:43
  • $\begingroup$ @LordSharktheUnknown -- Addition arises in even more contexts, but we always use $+$. $\endgroup$ – mr_e_man Sep 10 '18 at 19:14
  • $\begingroup$ $a^Tb$ is the expression of a particular inner product relative to an specific, but often tacitly-understood, basis. $\endgroup$ – amd Sep 10 '18 at 20:20

As you said, the notation $a\cdot b$ suggests bilinearity. The notations $\left<a,b\right>$ or $\left<a\mid b\right>$ or $\left(a,b\right)$ are better for non-bilinear scalar products, as those typically used in complex vector spaces, where you are as least as interested in sesquilinear forms that are linear in one component and antilinear in the other. For example, $$\left<u,\alpha v\right> = \alpha \left<u,v\right>, \quad \left<\alpha u,v\right> = \bar \alpha \left<u,v\right>, $$ where $\bar \alpha$ is the complex conjugate of $\alpha$. (Whether the scalar product should be linear in the left or in the right component is a question of taste and of whether you are a physicist).

  • $\begingroup$ One interesting solution is to use two-sided scalar multiplication: $cv=vc^*$. Now we can have it both ways: $$c(v\cdot w)=(cv)\cdot w=v\cdot(c^*\,w)$$ $$(v\cdot w)c=v\cdot(wc)=(vc^*)\cdot w$$ This is natural when described with a Clifford algebra (over Lorentzian $\mathbb R^{n,1}$) model of the complex inner product space. (There may be other real models.) $\endgroup$ – mr_e_man Sep 11 '18 at 14:34

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