I'm having trouble understanding how two D-branes leads to a $U(2)$ gauge theory from David Tong's notes, chapter 7, pages 191-192. I am learning group theory and I understand that a 'charge' is a generator of the symmetry transformation, in this case $U(1)$ of $A_{\mu}$. But now you have non Abelian gauge fields that are matrices and off-diagonal elements represent strings stretched between two different D-branes.

Here is what I do not understand:

  1. Under the $U(1)$ symmetries transformation, the fields have $(A_{\mu})^{1}_{2}$ and $(A_{\mu})^{2}_{1}$ have charges $(+1,-1)$ and $(-1,+1)$ respectively. How do we know that this is true and what does this mean?

  2. This implies that the gauge theory is $U(2)$. How does it follow from the above?


3 Answers 3


The answer to your second question is the easier one, because it is purely about the structure of the Lie algebra $\mathfrak{u}(N)$:

The group $\mathrm{U}(N)$ has dimension $N^2$ and its maximal torus is a subgroup $\mathrm{U}(1)^N$. If you split the algebra accordingly as $\mathfrak{u}(N) = \mathfrak{u}(1)^N \oplus \mathfrak{h}$, and write the generators of $\mathfrak{u}(1)^N$ as $T^{ii}, i\in\{1,\dots, N\}$, then $\mathfrak{h}$ consists of $N^2 - N$ generators $T^{ij}, i\neq j$ with $[T^{ii}, T^{jk}] = \delta^{ij}T^{jk} - \delta^{ik}T^{jk}$. The physicist says that $T^{jk}$ has "charge +1" under $T^{jj}$ and "charge -1" under $T^{kk}$.

So what needs to be established is that the massless states corresponding to the strings stretching from the first to the second D-brane really transform like that under the infinitesimal transformation generated by the $\mathrm{U}(1)$-symmetries associated with the respective brane. In the end, it will turn out that we really need to put it in by hand in some sense, but maybe it's at least an interesting history lesson:

You can't really see this if you take "$N$ coincident branes" as your starting point because "$N$ coincident branes" doesn't actually mean anything! What does it mean for branes to be coincident? A D-brane was originally just the surface to which Dirichlet boundary conditions confine the endpoint of an open string - there is no mathematical content to saying there is "more than one" such surface at the same point.

So why do physicists talk about these coincident branes? It's because of $T$-duality applied to an (ancient, in string theory terms) idea:

Originally, physicists simply associated the Chan-Paton factors of a $\mathrm{U}(N)$ gauge theory to the ends of strings ad hoc - Chan and Paton were "old" string theorists for whom the string was the flux tube between a quark and an antiquark, and they simply needed to put $\mathrm{U}(N)$ groups in there because they knew or at least suspected that quarks were charged under some $\mathrm{U}(N)$ groups.

Now, we come to Polchinski et al.'s "Notes on D-branes":

During the 90s, people started to think of D-branes as dynamical objects in their own right, and realized that the Dirichlet boundary conditions could be obtained by T-duality (one dimension of spacetime is compactified as a circle of radius $R$ and we send $R\to \alpha/R$) from Neumann boundary conditions - you get a D-brane sitting at one point in the compactified dimension. So a natural question is what happens to a string with ad hoc Chan-Paton factors when you dualize it into a string with Dirichlet conditions.

If you don't do anything special, nothing. There's nothing there hinting at some stack of branes. But if you break the symmetry as $\mathrm{U}(N)\to \mathrm{U}(1)^N$ prior to applying the duality, then suddenly you get not one $D$-brane, but $N$ $D$-branes, sitting around the circle at angles corresponding to the non-vanishing angle $\theta_i$ in the order parameter (value of a Wilson line) $\mathrm{diag}(\theta_1,\dots,\theta_N)$ of the spontaneous breaking of the Chan-Paton theory. The (massive) vector states associated with strings between two different branes are charged in the $(-1,1)$ fashion under the $\mathrm{U}(1)$ on these branes because we got them by dualizing a broken $\mathrm{U}(N)$ theory.

And now, if you take the limits $\theta_i\to 0$, then you see the D-branes in the dual theory rushing towards each other, until at $\theta_i = 0$, i.e. restored $\mathrm{U}(N)$ symmetry, there's only a single position left and they "all sit on top of each other". This is the origin of "coincident branes", and so by "$N$ coincident $D$-branes", we really mean "the theory T-dual to the theory of a string with Neumann boundary conditions with ad hoc $\mathrm{U}(N)$ Chan-Paton factors.


Ground states of strings stretching two $D$-branes are distinguished by two label integers that indicate the brane on which the open string endpoints lie according to the string orientation.

The sector $|p,^{+},p^{T};[12]\rangle$ parametrize strings with momentum labbels $p,^{+},p^{T}$ and embbeding functions $X^{\mu}(\tau,\sigma)$ with $\sigma \in [0,\pi]$ such that its $\sigma=0$ endpoint lie at the first brane and its $\sigma=\pi$ endpoint at the second one; similarly $|p,^{+},p^{T};[21]\rangle$ concern strings with the opposite orientation to the one defined by the aforementioned $[12]$ sector and finally, the sector $|p,^{+},p^{T};[ii]\rangle$ with $i = 1,2 $ do its own with strings whose both endpoints lie at the same $i$ brane.

From the latter construction you obtain four massless gauge bosons if the two branes are coincident, exactly the number of generators needed to realize the Lie algebra of $U(2)$.

How do you know that this configuration actually gives rise to a $U(2)$ gauge theory and not to a $U(1)^{2}$ one? That's because of string interactions. If you know why the wolrdvolume theory of a single brane is $U(1)$ and identify the sectors $[11]$ and $[22]$ as the photon generators of the $U(1)$ gauge theory on the branes 1 and 2 respectively, then would be able to recognize that the $[12]$ and $[21]$ massive sectors can't be embeeded into the $U(1)^{2}$ theory, that's because its massive character, and the fact that in perturbative string theory any sector $[ij]$ can interact with another $[jk]$ to produce a string from the sector $[ik]$ signal the possibility of a spontaneaous symmetry breaking ($SU(2) \rightarrow U(1) \times U(1)$) origin of the mass on the $[12]$ and $[21]$ massive sectors. Of course that the latter argument is not a proof, I'm just trying to give some intuition.

General case:

Everything becomes much more beaufiful and impressive if you reverse the logic. Consider the $SU(N)$ Lie algebra with generators $T^{a}$ and $T^{b}$ defined by the usual commutation relations $$[T^{a},T^{b}]=if^{abc}T^{c}.$$

Now recall the definition of the adjoint representation of the Lie algebra for $SU(N)$ as $$(T_{a})^{bc}=-if^{abc}.$$ It's very interesting to notice for a fixed index $a$ the generators $(T_{a})^{bc}$ correspond to a gluon in the $SU(N)$ gauge theory and that they are characterized by two free positive integer labbels, namely $b$ and $c$. It follows that in the general case you may think on $(T_{a})^{bc}$ as an open string in a theory of $N$ $Dp-$branes with edpoints at the $b$ and $c$ branes.

The beauty of string theory even allows you to geometrically engineer a lot of other wonderful gauge theory constructions, from gluons and quarks for all the simply laced Lie groups and an impressive plethora of nonperturbative properties otherwise impossible to visualize. Tong's lectures on solitons is an excellent reference to begin to learn how to achive this.

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    $\begingroup$ Saying that the number of generators matches sidesteps the actual question: How do you know that these $N^2$ generators generate $\mathrm{U}(N)$ and not e.g $\mathrm{U}(1)^{N^2}$? $\endgroup$
    – ACuriousMind
    Jun 28, 2020 at 18:59
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    $\begingroup$ Shouldn't the string with endpoints on the same D-brane have the Chan Paton factor | 1 1> and |2 2>? So is the idea basically that the number of gauge bosons in our theory equals the number of generators of the gauge group? And the four possible orientations of the string ([1,1], [1,2],[2,1] and [2,2]) give rise to the four elements of the gauge field and hence four gauge bosons? $\endgroup$
    – saad
    Jun 29, 2020 at 14:14
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    $\begingroup$ Of course you're right, strings with both endpoints lying at the same brane have $[11]$ and $[22]$ Chan-paton factors, thanks for being alert. Exactly, that's a wonderful summary of the idea of the answer. There are some subtleties that you should learn, but at this moment I humbly recommend you to not be too bothered about this. The most important one is that the worldvolume of $N$ coincident $D$-branes is really $SU(N)$, the reason for this is that a stack of $N$ build a gauge theory with group $U(N)$ as was discussed, but it's possible to prove (by using perturbative string theory) that ... $\endgroup$ Jun 29, 2020 at 15:24
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    $\begingroup$ $U(N)$ factorize as a semidirect product $SU(N) \ltimes U(1)$ and the $U(1)$ part decoples from the theory. For the $SU(2)$ case the dynamical Chan-Paton factors are $[11]$, $[22]$ and $\frac{1}{2}([12]+[21])$ correspond to the $W^{+}$, $W^{-}$ and the $Z^{0}$ gauge bosons respectively, the fourth linear combination $\frac{1}{2}([12]-[21])$ is also a singlet and (as I said) decoples. My opinion is that it's better to start by gaining intuition about how to visualize and engineer QFTs, the details should appear in due course. $\endgroup$ Jun 29, 2020 at 15:33
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    $\begingroup$ Other good and free references to learn computational details (besides the fantastic Tong's lectures) are: Joe's Little Book in String Theory and Introduction to Superstring Theory . Good luck. $\endgroup$ Jun 29, 2020 at 15:42

Tong gives a vanilla flavor of this in chapter 3 of those same notes. Start with a Dirac action and reparameterize the coordinate. If you can imagine a visual of two branes connected by opposite ends of a string one at sigma = 0, and the other at sigma = 2pi. You can do this for strings with many more configurations encoded by Chan-Paton factors.

The mass squared operator encodes the states.

It seems an interesting path forward might be to compute the Chan Paton factors for the U(2) or U(1) case and then the charge density and string charges. This should settle 1. and 2.


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