I am reading a paper that starts talking about 'left vector bundles' and I'm having trouble figuring out what they mean. The specific setup is as follows:

A quarternionic line bundle $L$ over manifold $M$ is a real smooth rank 4 vector bundle with fibers 1-dimensional quaternionic right vector spaces (*).

A complex quarternionic vector bundle is a pair $(L,J)$ with a quaternionic linear endomorphism $J$ such that $J^2=-1$.

A complex quarternionic vector bundle is thus a rank 2 left complex vector bundle whose complex structure is compatible with the right quaternionic structure (**).

(*) Is a right vector space a vector space $V$ with scalar multiplication only defined on the right? So $V \times F \to V$ for field $F$ but $F \times V \to V$ is not defined?

(**) If what I say above is correct, what does the compatibility mean? If this complex quaternionic vector bundle is left, then the compatability means

$a(QJ)=(aQ)J,~a\in F, Q\in \mathbb{H}$

or something? Or does the "rank 2" say the compatability is something like

$\mathbb{H}\oplus J\mathbb{H}=\mathbb{H}J\oplus \mathbb{H}$

The paper I am reading is "Quaternionic Analysis on Riemann Surfaces and Differential Geometry" by Pedit and Pinkall (1998). It seems like my confusion is not related at all to the quaternion structure, would apply to any vector space. Also, I have found some other references to the left- and right- vector bundles in double vector bundles, but that also seems not related to this. Does anyone have any clarity?

  • $\begingroup$ For a vector space a-priori you always have only multiplication with scalars from the left. Of course defining right multiplication just as $v\cdot \lambda := \lambda \cdot v$ does not give any deeper theory. Also I have not yet come across where a different right multiplication has been defined. (This would lead to confusion and you could just as well define a different left-action of your field.) $\endgroup$ – bodo Jun 29 '12 at 19:13

(*) Is a right vector space a vector space V with scalar multiplication only defined on the right? So V×F→V for field F but F×V→V is not defined?

This is correct. The main point is the associativity rule: if $v$ is a vector and $a,b$ scalars, then in a left-vector space we have

$$ a \cdot (b \cdot v) = (a \cdot b) \cdot v$$

and in a right-vector space we have

$$ v \cdot (a \cdot b) = (v \cdot a) \cdot b.$$

One could always just swap the order of the factors to write scalar multiplication on the left in a right-vector space, but that is potentially very confusing, because the associativity rule would be

$$\color{red}{ b \cdot (a \cdot v) = (a \cdot b) \cdot v.}$$

I've colored this equation red because it's a bad idea! Among other things, it would mean that we are not allowed to write $abv$, because the two different interpretations $(ab)v$ and $a(bv)$ can be different.

However, for a commutative field (like the complex numbers), $ab = ba$ so we don't have to develop separate notions of left and right vector spaces in that context.

It is possible to talk about a left-$F$ right-$E$ vector space, where $F$ and $E$ are skew fields: this is a vector space that is a left $F$-vector space and a right $E$-vector space that are "compatible": they satisfy an additional associativity constraint:

$$ f \cdot (v \cdot e) = (f \cdot v) \cdot e. $$

Unfortunately, I'm not familiar with your context, so I can't answer your questions directly. In fact, I don't think I've ever done linear algebra over skew fields before -- all of these ideas I'm familiar with from module theory. But, at least, module theory is a generalization: a left vector space over a (skew) field $F$ is the same thing as a left module over $F$ (and the same on the right), so I'm assuming all of the notions you're talking about have the same meaning as the module-theoretic version.

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    $\begingroup$ I think part of the reason for confusion is that a "(left or right) quaternionic vector space'' is not really a vector space at all -- it's a (left or right) module over the quaternions. At least, every definition of vector space that I've seen requires the set of scalars to be a field, which the quaternions are not. A "quaternionic vector bundle'' is a real vector bundle and also a bundle of left- or right- modules over the quaternions. $\endgroup$ – Jack Lee Jun 29 '12 at 22:03
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    $\begingroup$ I think it is not too harmful to extend "vector space" to mean "left module over a division ring." Most of the theorems look the same. $\endgroup$ – Qiaochu Yuan Jun 29 '12 at 22:14
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    $\begingroup$ @Jack: While I've not studied it in any depth, I have definitely seen texts that define vector spaces over any skew field, not merely fields, and my impression had always been that it's standard enough that one shouldn't bat an eye at it, but uncommon enough that it deserves explicit mention when you're discussing the non-commutative case. In fact, I made a deliberate choice to use the phrase "skew field" instead of my preferred "division ring" to emphasize this. $\endgroup$ – user14972 Jun 30 '12 at 1:36

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