Is $U(1)\times SU(2) \times SU(3)$ a vector space over a field? I saw an article here that seemed to me that a similar concept to a field extension was being used.

In QFT, is each particle considered to be its own vector space?

And are the individual vector spaces then bridged together by these group extensions?

If I have asked too many questions, feel free to respond to just one.

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    $\begingroup$ Note that in the literature one often finds authors talking about a Lie group when they really mean the corresponding Lie algebra, and vice-versa. A Lie group is in general not a vector space, while a Lie algebra is always a vector space. $\endgroup$
    – Qmechanic
    Commented Sep 3, 2013 at 15:00
  • $\begingroup$ @Qmechanic: Aaah, thanks Qmechanic.... very much!!! I appreciate that! $\endgroup$ Commented Sep 3, 2013 at 15:45

2 Answers 2


Since I have gleaned some more idea of "where you are" in your learning and shown considerable enthusiasm for getting a thorough grasp of fundamentals, I'd like to add some more details (from a non-particle physicist, mind you, so there are many aspects of your question I must steer clear of) to Lubos's excellent answer. Also, in your question you spoke of extension fields and analogies with Lie groups and this suggests to me that you are thinking about analogies between Lie theory and Galois theory, so I shall touch on that idea too: there are indeed likenesses and it was precisely Lie's wish for a "Galois" theory of continuous groups that lead him to found Lie theory.

What the Lie Bracket remembers about the group: The Baker Campbell Hausdorff Theorem

Let us finalise Lubos's excellent intuitive statement "the curvature of the sphere (Lie group) is fully remembered by the commutator operation on the Lie algebra, however". Another variation on this memorable statement is that the "Lie algebra encodes almost all of the information about the Lie group". If you remember these you won't go too far wrong. I'd slightly disagree with Lubos's "fully remembered" but this is a very near miss so here are some final pieces to the jigsaw; you can see that something else is needed since two different Lie groups can have precisely the same Lie algebra, for example the pair $SU(2)$ and $SO(3)$ as well as (a different example) the pair $U(1)$, which is compact, and the noncompact $(\mathbb{R},\,+)$ (this latter one isomorphic to $(\mathbb{R}^+\sim\{0\},\,\times)$).

Lubos's statement (Lubos, correct me if this is not a good representation of your words) is encoded in the Baker Campbell Hausdorff Theorem. The exponential maps neighbourhoods of the origin in the Lie algebra vector space (call it $\mathfrak{g}$) one to one onto neighbourhoods of the identity in the Lie group $\mathfrak{G}$: the locally well defined logarithm does the inverse. Then it can be shown that there is a neighbourhood $\mathbf{N}\subset\mathfrak{g}$ (it has to be "small enough" as defined by a suitable metric in $\mathfrak{g}$) such that for $X,\,Y\in\mathbf{N}\subset\mathfrak{g}$ (and so $e^A,\,e^B\in \mathfrak{G}$) there is a $Z\in\mathfrak{g}$ such that $e^X\,e^Y = e^Z\in \mathfrak{G}$ and:

$Z = X + Y + \frac{1}{2} [X,\,Y] + \frac{1}{12}[X,\,[X,\,Y]] - \frac{1}{12}[X,\,[Y,\,X]] + ...$

where ALL the terms involve only the Lie bracket (commutator). The exact co-efficients in this formula are extremely complicated to write down (there is a fearsome formula owing to Dynkin, see [Rossmann] in chapter 1 (I give references at the end)), but their exact values are not important for this discussion. What is important is that the BCH formula only involves Lie brackets, sums and scalar multiplications, so partial sums yield nothing outside the Lie algebra; moreover it is convergent in a suitably small neighbourhood of the zero vector, so it must converge to a Lie algebra member (Lie algebras, also being vector spaces as you know, are closed, hence the limit is in the Lie algebra). So the BCH formula "pulls the group multiplication back through the exponential function" and shows that in small enough neighbourhoods of the identity / origin of the Lie group / algebra, the group multiplication is indeed wholly and exactly defined by the Lie bracket.

What the Lie Bracket does not "remember" about the group: Global Topology and the Fundamental Group

But there is still some slight "wriggle room" (ambiguity: there is more than one way to consistently define the exponential) in the definition of the nonlinear $\exp$ mapping the algebra to group - equivalently - ambiguity in the definition of the group multiplication. To understand this, witness the two different Rodrigues formulas mapping the same Lie algebra $su(2) \cong so(3)$ to the topologically different $SU(2)$ and $SO(3)$:

$$\begin{array}{rll} su(2)\mapsto SU(2):& H_{2\times 2}\mapsto\exp\left(H_{2\times 2}\right)=\cos\left(||H_{2\times 2}||\right)\;I_{2\times 2} +\frac{\sin\left(||H_{2\times 2}||\right)}{||H_{2\times 2}||}\, H_{2\times 2}\\ so(3)\mapsto SO(3):& H_{3\times 3}\mapsto\exp\left(H_{3\times 3}\right)=I_{3\times3}+\frac{\sin\left(||H_{3\times 3}||\right)}{||H_{3\times 3}||}\,H_{3\times 3} +\frac{1-\cos\left(||H_{3\times 3}||\right)}{||H_{3\times 3}||^2}\,H_{3\times 3}^2\\ \end{array}$$


$$H_{2\times2} = \left(\begin{array}{cc}i\,z&i\,x+y\\i\,x- y&-i\,z\end{array}\right)\quad\quad H_{3\times3} = \left(\begin{array}{ccc}0&z&-y\\-z&0&x\\y&-x&0\end{array}\right)$$

and $||H_{2\times2}||=||H_{3\times3}||=\sqrt{x^2+y^2+y^2}$. (In $SO(3)$ the BCH formula has a "closed form" expression, see [Engø], and a like closed form expression for $SU(2)$ follows by the same "tricks"). The same exponential Taylor series is used in each case, it's simply that the nonlinearity manifests itself differently in each case owing to the Cayley-Hamilton theorem's working on different characteristic equations fulfilled by $H_{2\times2}$ and $H_{3\times3}$.

The final ingredient in all of this, as discovered in 1925 by Otto Schreier, is the Lie group's global topology (see [Stillwell] chapter 8) and this information is encoded in a group's:

  • Fundamental group, which is the discrete group of all homotopy classes defined by loops passing through the group's identity. Strictly speaking, one must specify different fundamental groups at each point within the group, but in a connected Lie group (or any other connected manifold) they are all isomorphic. So we have the result that completes Lubos's memorable statement in the connected case:

A connected Lie group is wholly specified by the Lie algebra (the "memory" of the commutators) together with "the" fundamental group

  • Discrete group of nonintersecting disconnected cosets are needed further to the above to wholly specify a Lie group with distinct nontrivial connected components: for example, the identity connected component $SO^+(1,3)$ of the Lorentz group (the "proper, othochronous" transformations that keep the orientation of space and direction of time) is the a normal subgroup of the whole Lorentz group $O(1,3)$ and the discrete group of cosets $O(1,3)/SO^+(1,3)$ is the Klein four-group $V_4$ (comprising $I, P, T$ and $P\cdot T$ where $P$ is the space inversion matrix $\operatorname{diag}(1,-1,-1,-1)$ and $T$ the time reversal matrix $\operatorname{diag}(-1,1,1,1)$).

To illustrate this global topology for our $SO(3)$, $SU(2)$ example: $SO(3)$ is not simply connected as hopefully you can understand by looking at my rough drawing below:


Here you are to imagine all the rotation operators in $SO(3)$ as points in a compactified Euclidean space: this could be thought of as a compactified version of the Lie algebra but don't get too caught up in it's being related to the algebra: the drawing is of a sphere of radius $\pi$ but it is a special sphere, where antipodean point pairs on its surface are "identified" - thought of as being the same point. To draw a rotation of angle $\theta$ about an axis defined by the unit vector $(\gamma_x,\,\gamma_y,\,\gamma_z)$, we first restrict angles to lie in the interval $(-\pi, \pi]$ so we get $\theta\mapsto\theta^\prime = \theta + 2\,k\,\pi\in(-\pi, \pi]$ by chopping off unimportant chuncks of whole number $k$ multiples of $2\pi$, then we draw a vector of length $\theta^\prime$ with its tail at the origin (the group's identity) and in the direction of $(\gamma_x,\,\gamma_y,\,\gamma_z)$: the point at the vector's head uniquely represents any element in $SO(3)$. Now we think about the fundamental group of $SO(3)$; there is the homotopy class of paths like $\Gamma$ which can be continuously shrunken to a point and those like $\Omega$ which cannot. Imagine riding on the $C^1$ path $\Omega$ through the Lie group: when we reach the point $P$ and go a tiny bit further, we find ourselves on the sphere's "opposite side", just beyond the point $P^\prime$ diametrically opposite to $P$. We keep going on this path $\Omega$ until we loop back to the identity. It should be clear that a loop like $\Omega$ cannot be continuously shrunken back to a point at the identity: because the path emerges from an antipodean point (actually the same point in our definition) as soon as it crosses the sphere's surface we need to "pull the path back though $P$" so we can loop back to origin, but we cannot do this as the path is joined beyond $P^\prime$ to the identity, so the homotopy class of $\Omega$ is an element of the fundamental group $\pi_1(SO(3))$ of $SO(3)$ distinct from the identity and so $\pi_1(SO(3))$ is not trivial. However, it should be fairly obvious to see a homotopy between $\Omega$ and its inverse loop (i.e. $\Omega$ run in the opposite sense): simply rotate this path through $180^o$ in my drawing above about the origin (i.e. the identity $I$). So this loop is not like winding a loop through a torus: we can continuously deform $\Omega$ into its inverse, whereas there is no way to make the arrows point the other way on a loop with arrows drawn on it threaded through a torus without first breaking the loop. So our fundamental group presentation is $\pi_1(SO(3))=\left<\Omega\,|\, \Omega^2=1\right>\cong\mathbb{Z}_2$.

We can form the universal cover (see the Wiki page on Covering Groups) of $SO(3)$ to find a simply connected group (there is a standard construction detailed on the Wiki page for doing this) and in this case we get the simply connected $SU(2)$ as the universal covering group. If $\mathfrak{G}$ is a connected Lie group with universal cover $\tilde{\mathfrak{G}}$, then the fundamental group $\pi_1(\mathfrak{G})$ is given by the quotient group $\pi_1(\mathfrak{G})\cong \tilde{\mathfrak{G}}/\mathfrak{G}$: in this case the quotient group is $\mathbb{Z}_2$, comprising the homotopy class of loops like $\Gamma$ in the drawing above (the identity of $\pi_1(SO(3))$) and the class of loops like $\Omega$. Another way to visualize this is through the very neat "topologist's belt trick" (sometimes call Dirac's belt trick) but I'm banging on a bit so you'd better look that one up - this one is very fun to demonstrate to smallish children about the age of seven or older as I've found it evokes a strong sense of wonder in them. Both $\omega\in SU(2)$ and $-\omega\in SU(2)$ form a coset in $SU(2)$ that is mapped to the same element of $SO(3)$ by the standard homomorphism that recovers rotation matrices from $SU(2)$ elements. The fundamental group for a Lie group (indeed any topological group) is always Abelian, so that a Lie group is a very special, restricted kind of manifold (manifolds in general can have any finitely generated free group as their fundamental groups). The universal cover has a discrete centre $\mathcal{Z}(\tilde{\mathfrak{G}})$ (=supbroup of elements that commute with all elements of $\tilde{\mathfrak{G}}$) and the set of all distinct possible Lie groups with the same Lie algebra $\mathfrak{g} = \operatorname{Lie}(\mathfrak{G}) = \operatorname{Lie}(\tilde{\mathfrak{G}})$ is in one-to-one, onto correspondence with the subgroups of the centre $\mathcal{Z}(\tilde{\mathfrak{G}})$ of the universal covering group: we count the trivial group and the whole of $\mathcal{Z}(\tilde{\mathfrak{G}})$ here and the adjoint representation (see the Wiki page with this name) of $\mathfrak{G}$ corresponds to the trivial subgroup of $\mathcal{Z}(\tilde{\mathfrak{G}})$ and is the farthest possible from being simply connected, and the universal cover group corresponds to the whole of $\mathcal{Z}(\tilde{\mathfrak{G}})$. These possibilities exhaust all possible Lie groups with the same Lie algebra. $SO(3)$ is the adjoint representation whereby $SU(2)$ acts on its Lie algebra $su(2)$ and $SO(3)$ is its own adjoint representation. The adjoint representation annihilates the group's centre, which is the kernel of the representation. This is why the BCH formula doesn't quite encode all the information about the group: whilst it can "see" a continuous part of the centre $\mathcal{Z}(\tilde{\mathfrak{G}})$ through the Euclidean sum $X + Y$ in the BCH formula, the higher order terms cannot see the discrete centre.

You should be able now to think of other example of $\mathfrak{G}=U(1)$ and $\tilde{\mathfrak{G}}=(\mathbb{R},\,+)$ in these terms: they both have the same Lie algebra $\mathfrak{g}=(\mathbb{R},\,+)$ and $\tilde{\mathfrak{G}}=(\mathbb{R},\,+)$ is the universal covering group of $\mathfrak{G}=U(1)$. The fundamental group of course is $\mathbb{Z} = \cdots ,\,-2,\,-1,\,0,\,1,\,2,\,\cdots$, where the integer $n$ corresponds to a loop around the circle $U(1) = \{e^{i\,\theta}:\;\theta\in\mathbb{R}\}$ comprising $n$ bights in the anticlockwise direction.

To help your intuition further: Lie groups are "almost always" matrix groups as follows. There is a corollary to a difficult theorem known as Ado's theorem that every Lie algebra can be realized as a Lie algebra of square matrices. The same is not true of Lie groups: not every Lie group can be represented as a group of matrices but it is almost true (a consequence of the Peter-Weyl theorem is that every compact group can be realized as a group of square matrices). Certainly, since we can find a square matrix realization for every Lie algebra, we can build a matrix Lie group with that algebra as its Lie algebra through the matrix exponential function; then we find that matrix group's universal cover and this is where we sometimes fail to get a matrix group. This is not typical and the first Lie groups that were not also matrix groups (so called metaplectic groups) weren't found until 1937. These oddballs are all covering groups of noncompact groups.

One should note parenthetically that the Lie algebra $\operatorname{Lie}(U(1)\times SU(2)\times SU(3))$ of the direct product $U(1)\times SU(2)\times SU(3)$ is the direct sum of the respective Lie algebras, so this is a result something like that which you may have been thinking of when you asked your question: in symbols:

$$\begin{array}{lcl}\operatorname{Lie}(U(1)\times SU(2)\times SU(3)) &=& \operatorname{Lie}(U(1))\oplus \operatorname{Lie}(SU(2)) \oplus \operatorname{Lie}(SU(3))\\ &=& u(1) \oplus su(2)\oplus su(3)\end{array}$$

Lie Groups and Galois Theory

Lie indeed imagined a Galois theory for continuous groups and there are analogies, but Lie theory is more complicated. Linear Lie algebras simplify the study of some properties of the nonlinear Lie group and there is a one-to-one correspondence between the Lie subgroups of a Lie group $\mathfrak{G}$ and Lie subalgebras of its Lie algebra $\mathfrak{g} = \operatorname{Lie}(\mathfrak{G})$: this is the so-called "Lie Correspondence" detailed in chapter 2 of [Rossmann] (the correspondence is discussed in section 2.5) just as the normal subgroups of a field extension's Galois group correspond to one to one with all the extension fields contained within the particular extension, and we can therefore study extension fields by studying the Galois group of automorphisms on them. In the other direction, the Lie group acts on its own Lie algebra through its own adjoint representation referred to above, just as the Galois group of automorphisms act on the extension fields they are used to study.

References and getting more info

There are three excellent references I would recommend:

Wulf Rossmann, "Lie Groups, An Introduction Through Linear Groups"

John Stillwell, "Naïve Lie Theory"

Brian Hall, "Lie Groups, Lie Algebras, and Representations: An Elementary Introduction"

Read the first two chapters of Rossmann and all of Stillwell thoroughly for fundamentals and the Hall book rushes the fundamentals a bit to get to representation theory, which is of most use to physicists. Stillwell doesn't cover representation theory - it is meant mainly for undergraduate level but is very worth reading. Stillwell's is the best description, in my opinion, of the global topology ideas discussed here.

Another remarkable little book

J Frank Adams "Lectures on Lie Groups"

shows how far one can get in studying the highly nonlinear Lie group without using the Lie algebra. It certainly surprised me.

There is also the paper I cited:

K. Engø, "On the BCH-Formula in SO(3)", BIT Numerical Mathematics 41 (2001), no.3, pp629--632.

As we saw, the Lie group's fundamental group is always Abelian (as is indeed the fundamental group of any topological group) and this is the reason why I'm not greatly fond of the modern approach that introduces the Lie group as a manifold with group structure: a manifold is far too broad and general a thing and you do not need anything like the whole of differential geometry to understand a Lie group well. Although it is good to abstract and generalize, you might say that this approach is viewing the trees of the forest from too far away for a good first look. Spivak's "Comprehensive Introduction to Differential Geometry" teaches Lie groups like this, whereas I intellectually went the other way using the [Rossmann], [Stillwell] and [Hall] references and then I used the intuition about Lie groups to help me understand the differential geometry of more general Riemannian manifolds: one begins with a simple set of axioms about the neighbourhood of the identity in the Lie group and $C^1$ paths therein and then defines the whole connected component as the smallest group containing this neighbourhood: in this way the Lie group's analytic manifold "builds itself" and indeed Montgomery, Gleason and Zippin's solution to Hilbert's fifth problem shows that not even differentiability needs to be assumed, for it emerges naturally from only the assumptions about a Lie group's continuity. The Lie group idea emerges from even more primitive assumptions in the case of compact semisimple Lie groups: for there is no other abstract group structure possible for such a Lie group so even the topology emerges from the algebraic structure alone and every group automorphism as an abstract group preserves the Lie group structure as well (van der Waerden, B. L., Mathematische Zeitschrift 36 pp780 - 786). Lie groups are very special indeed and the modern idea of a manifold contains too much machinery to see them clearly with. Intuitively, this highly special nature comes from homogeneity- the fact that the group action clones the around-identity-neighbourhood structure throughout the whole manifold and there just aren't that many axiom systems and behaviours that can withstand such wholesale "cloning" and still be consistent.

  • $\begingroup$ Here's my first comment: WOW!!!!!!! $\endgroup$ Commented Sep 4, 2013 at 21:25
  • $\begingroup$ This is an incredible answer... I don't know how I can repay you for sharing this quality and breadth of information with me.... $\endgroup$ Commented Sep 4, 2013 at 21:37
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    $\begingroup$ @curiousGeorge119 BTW don't be put off by my comment "harder than it should be": I'm not meaning to sound like it's all trivial for anyone with any brains - it's not - it takes considerable learning effort. I just think the quality of technical writing around could be better. This is a subject where you will need some care and practice - indeed I once wrote software for computing the Dynkin co-efficients in the BCH series and I found my brain just didn't compute all the nested Lie brackets - it was quite a surreal feeling to feel one's mind totally failing to grasp a concept and I had to ... $\endgroup$ Commented Sep 5, 2013 at 13:13
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    $\begingroup$ @curiousGeorge119 You might like to check out my demo "Dirac Belt Trick Simulation Showing Double Cover of SO(3) by SU(2)" at Wolfram Demonstrations if you've not met the universal cover before, a good lecture on it is docstoc.com/docs/28157208/… . If you think about the abstract procedure, you can see that the ribbon in the belt trick encodes a path from the identity to the SO(3) transformation encoded by ..... $\endgroup$ Commented Nov 4, 2013 at 0:54
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    $\begingroup$ @curiousGeorge119 .... the doll's orientation in space. Every deformation of the ribbon thus encodes a member of the same homotopy class, so the ribbon itself pretty much encodes THE homotopy class (of course there will be some deformations that the ribbon's elasticity and physical properties will not let it reach, but its an extremely close analogy). In particular, looping the ribbon over the doll with the latter held fixed leads to a path in the same homotopy class: so if you can undo the twist by looping, the doll and ribbon still encode the same member of $SO(3)^\sim \cong SU(2)$. $\endgroup$ Commented Nov 4, 2013 at 0:58

No, the group $U(1)\times SU(2)\times SU(3)$ isn't a vector space of any kind because it doesn't have any (commuting) addition operation (curved group manifolds can rarely have such a structure).

The article "group extension" you linked to makes it very clear that the group extension does not have to be a vector space and the group operation does not have to be Abelian.

Field extensions are commutative (and vector spaces) because a field itself is a commutative ring. But the Standard Model group isn't based on any field in this mathematical sense and the group is non-Abelian i.e. non-commuting – that's what physicists describe as "groups in Yang-Mills theory".

No, it is not true that individual particles "are" vector spaces. QFT, like any quantum theory, has an important complex vector space, the Hilbert space. No other spaces appearing in the formalism of QFT are vector spaces in general. This also answers your last question by negating its assumptions.

Just to be sure, one may consider "first-quantized" theories or one-particle sectors of QFTs. They have their own Hilbert space you could interpret as the vector space for "one particle". But you could have meant something completely different; it wasn't clear what role your hypothetical vector space connected with one particle should play.

  • $\begingroup$ Wow, that first statement you make is excellent. $\endgroup$ Commented Sep 3, 2013 at 13:55
  • $\begingroup$ Thank you for your answer, it is fantastic. "Field extensions are commutative because a field itself is a commutative ring". That is absolutely right!! Although the Lie Algebra is closed under commutative multiplication, commutative addition is not, is that correct? $\endgroup$ Commented Sep 3, 2013 at 14:00
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    $\begingroup$ @curiousGeorge119 Your last question is a bit wonky: a Lie algebra $\mathfrak{g}$ over a field $K$ is closed as follows under all these three operations: (i) $a\in K, X\in \mathfrak{g}\Rightarrow a\,X\in \mathfrak{g}$; (ii) $X,\,Y\in \mathfrak{g}\Rightarrow X+Y\in\mathfrak{g}$; (iii) $X,\,Y\in \mathfrak{g}\Rightarrow [X,\,Y]\in\mathfrak{g}$; so, in particular, $\mathfrak{g}$ is closed under addition as well as under scalar multiplication (which you seem to grasp). Does this answer your question? $\endgroup$ Commented Sep 3, 2013 at 14:21
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    $\begingroup$ @WetSavannaAnimal aka Rod Vance: Thanks for your comment... so it appears a Lie Algebra is a vector space, but a Lie Group is not... $\endgroup$ Commented Sep 3, 2013 at 15:48
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    $\begingroup$ Thanks for your excitement, Curious George. And yes, exactly, a Lie algebra is a vector space but a Lie group is not (except for $U(1)$ in some trivial sense). A Lie algebra is a vector space that may be imagined or defined as the tangent space of the group manifold (imagine a three-sphere for $SU(2)$) attached at the identity (North pole). The tangent space is clearly a linear vector space but the sphere is curved and therefore "nonlinear". The local character of the curvature of the sphere (Lie group) is fully remembered by the commutator operation on the Lie algebra, however. $\endgroup$ Commented Sep 3, 2013 at 17:32

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