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Let $V$ denote an inner product space. Write $V^*$ for either the algebraic dual, or else the continuous dual. In either case, for each vector $v \in V$, we get a covector $v^c \in V^*$ given by:

$$v^c = \langle v,-\rangle$$

Question. We call $v^c$ the [what] of $v$?

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    $\begingroup$ We call $v^c$ the dual of $v$. If $v^c$ actually the function that maps $w$ to $\langle v,w\rangle$. $\endgroup$ – Lærne Jul 22 '15 at 14:06
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    $\begingroup$ In quantum mechanics it is called bra-v. $\endgroup$ – A.Γ. Jul 27 '15 at 21:35
  • $\begingroup$ The Duke of Earl. youtube.com/watch?v=j9PoUsRibtE $\endgroup$ – Will Jagy Jul 27 '15 at 22:56
  • $\begingroup$ One can also say that $v^c$ is the covector produced by lowering the index of $v$ (with the inner product). $\endgroup$ – Travis Willse Jul 28 '15 at 20:07
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$\renewcommand{ket}[1]{|#1\rangle}$ $\renewcommand{bra}[1]{\langle#1|}$ There are several common terminologies:

  • As you might guess based on the fact that $V^*$ is called the "dual" of $V$, in mathematics and mathematical physics the covector of a vector $v$ is called the "dual" of $v$.

  • In physics, vectors are often denoted e.g. $\ket{v}$. The $v$ is a label which names the vector. In other words, $\ket{v}$ is like $\vec{v}$. The associated covector is denoted $\bra{v}$. The notation is justified by the fact that if you line up the covector and the vector you get $$ \bra{v}v\rangle$$ which is the usual symbol for an inner product. The symbol $\ket{\cdot}$ is called a "ket" and the $\bra{\cdot}$ is called a "bra" so that when you line them up to make an inner product you get "braket". In this system, the association between the vector and associated covector is expressed by using the same label, $v$, in either the ket or the bra symbol. In speech we just call $\bra{v}$ "bra $v$", as opposed to "ket $v$". I think many people would still say "$\bra{v}$ is the dual of $\ket{v}$".

  • You mentioned inner product spaces. In fact, the inner product function $\bra \cdot \cdot \rangle$ can be thought of as a rank-2 tensor: when you feed it two vectors it spits out a scalar. This tensor can be denoted in index notation as $g^{\alpha \beta}$. The inner product of two vectors $U_\alpha$ and $V_\beta$ is written $$\bra{U}V\rangle = g^{\alpha\beta} U_\alpha V_\beta \tag{1}$$ where matching indices indicates contraction. The upper indices are called "covector" indices because they indicate the need to consume a vector in order to produce a scalar.$^{[a]}$ So, if we have something like this $$U^\alpha \, ,$$ that thing must be a covector because it contracts with a single vector to yield a scalar. Therefore, upper indices indicate covector parts of the tensor and lower indices indicate vector parts. Now, if we contract $g$ with only one vector $$ g^{\alpha \beta}V_\beta$$ we're left with a free upper index, i.e. a covector. This thing is usually denoted $V^\alpha$ and is precisely the covector associated to $V_\beta$, a.k.a the dual of $V_\beta$. In this system you might also call $V^\alpha$ the "covariant version of $V_\beta$" or say that $V^\alpha$ is "$V_\beta$ with its index raised".

The point of this answer is to list the various common ways that the covector versions of vectors are called, and provide some motivation for the terminology.

$[a]$: It's also common to call upper indices "covariant" and lower indices "contravariant". This has to do with how the components of representations of vectors and covectors, expressed in a given basis, transform under coordinate transformations.

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  • $\begingroup$ The OP has requested specific references. I personally was unable to find any reputable source for the first one you listed ("dual" of $v$), even though I consider it the right answer. Do you have a reference for it? $\endgroup$ – Zev Chonoles Jul 28 '15 at 8:17
  • $\begingroup$ @ZevChonoles I don't see a request for references from the OP anywhere on this page. If I ctl-f the page for "Goblin" I see only the original post and the name showing up in the bounty note. $\endgroup$ – DanielSank Jul 28 '15 at 9:03
  • $\begingroup$ Look at the bounty posting. $\endgroup$ – Zev Chonoles Jul 28 '15 at 9:06
  • $\begingroup$ @ZevChonoles good point! My sources here are really just 11 years of physics/math education and experience. I'd hesitate to ever say that notation and terminology are "official" because, as illustrated in my answer, those things vary between different groups and fields. $\endgroup$ – DanielSank Jul 28 '15 at 9:55
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Echoing @Lærne, $v^c$ is the dual of $v$. If you want a reference, how about Halmos's "Finite-Dimensional Vector Spaces"? In the second edition, sections 13, 14, and 15 along with sections 67, 68, and 69 may be what you're looking for.

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  • $\begingroup$ I don't see an introduction of that terminology anywhere in the sections you cited. Could you give a specific page reference? (But note that I completely agree that "dual" is the correct word, I'm only asking about your reference.) $\endgroup$ – Zev Chonoles Jul 28 '15 at 8:14
  • $\begingroup$ @ZevChonoles Honestly, I've never seen using the term "dual" to the vector from Little Riesz theorem. It sounds a bit technical (applied science terminology?). Dual basis is used, but it is different. If I google it most links give "dual vector space" (which is not the space of dual vectors). $\endgroup$ – A.Γ. Jul 28 '15 at 10:49

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