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Semi-stable quivers in susy gauge theories. An example I do not understand.

+ 4 like - 0 dislike
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Let us consider the the quiver with 2 nodes and 1 arrow (we can think of some specific physical system, e.g. two points of the moduli space of the Coulomb branch of a $\mathcal{N}=2$ theory). Then, we can write this as $1 \to 2$. Now let us consider some representations of this quivers. 

  1. $F_0=(\{0,0\}, 0)$
  2. $F_1=(\{ \mathbb{C},0\}, 0)$
  3. $F_2=(\{0, \mathbb{C} \}, 0)$
  4. $F_3=(\{ \mathbb{C}, \mathbb{C} \}, 0)$
  5. $F_4=(\{ \mathbb{C}, \mathbb{C} \}, z)$

where $z\in \mathbb{C}$. Of course the notation $(\{ A, B\}, x)$ means that the vector space of the node 1 is $A$ the vector space of node 2 is $B$ and the map between them is $x$. Ok, now, using King's parameters $(\theta_1, \theta_2)$ the slope is defined as 

$$ \mu(F_i) = \frac{ \theta_1 d_1 + \theta_2 d_2 }{ d_1 + d_2} $$

where $d_i$ is the dimension of the vector space of node $i$. Now a representation is called $semi-stable$ for $(\theta_1,\theta_2)$ if for any sub representation $F' \subset F$ we have $\mu(F') \leq \mu(F)$. Now, there exists a second stability condition arising from extended supersymmetric gauge theories. There, we have a central charge function. For the specific quiver it is a map

$$ Z : \mathbb{Z}^{2} \to \mathbb{C} $$

such that the central charges associated to the nodes $Z(d_1,d_2)$ lie in a half-plane $H_{\phi}$ defined by

$$ H_{\phi} = \{ z \in \mathbb{C}^{*} | \, \phi < \text{Arg}(z) \leq \phi + \pi   \} $$

Then a representation is called semi-stable if $\text{Arg}(Z(d_1(F'), d_2(F'))) \leq \text{Arg}(Z(d_1(F), d_2(F))) $. Let me mention I am already a bit lost here, with this definition of semi-stability.

Now it is possible to choose King's parameters as a function of the central charge as following

$$ \theta_1 =  \frac{Z(\delta_1)}{Z(d_1)}, \,\,\,\,\,\,\,   \theta_2 =  \frac{Z(\delta_2)}{Z(d_2)}$$

where $\vec{\delta}_i = (0, \ldots, 1 , \ldots, 0) $, i.e. 1 at position i.  In our case, with 2 nodes of course we only have the vectors $\delta_1 = (1,0)$ and $\delta_2 = (0,1)$. Now, and this is my second and bigger confusion is how can we determine from the above that for $\theta_1 < \theta_2$ the semi-stable representations are $F_0, F_1, F_2$ while for $\theta_1 > \theta_2 $ the semi-stable representations are $F_0,F_1,F_2,F_4$. How do we get this?

This is taken from this set of notes pages 3,4,5.

asked Jul 1, 2015 in Theoretical Physics by conformal_gk (3,535 points) [ revision history ]
edited Jul 1, 2015 by 40227

1 Answer

+ 3 like - 0 dislike

It is enough to check the definition. As $F_0$, $F_1$ and $F_2$ have no non-trivial subrepresentations, they are always semistable, whatever the values of $\theta_1$ and $\theta_2$ are.

One has $\mu(F_1)=\theta_1$, $\mu(F_2)=\theta_2$, $\mu(F_3)=\mu(F_4)=(\theta_1+\theta_2)/2$.

As $F_1$ and $F_2$ are subrepresentations of $F_3$ and as $\mu(F_3)-\mu(F_1)=(\theta_2-\theta_1)/2$ and $\mu(F_3)-\mu(F_2)=(\theta_1-\theta_2)/2$, we see that $F_3$ if not semistable if $\theta_1<\theta_2$ or $\theta_2<\theta_1$.

As $F_2$ is the only non-trivial subrepresentation of $F_4$ and as $\mu(F_4)-\mu(F_2)=(\theta_1-\theta_2)/2$, $F_4$ is semistable for $\theta_1 >\theta_2$ and non semistable for $\theta_2 > \theta_1$.

EDIT (taking into account the comment). The case of $F_0$, i.e. of the zero representation is a bit degenerate because the slope is not well defined for it. I think the usual convention/definition is to say that it is semistable because it has no non-tivial subrepresentation.

Let me explain more precisely why $F_2$ is a subrepresentation of $F_4$. The spaces ${0,\mathbb{C}}$ composing $F_2$ are naturally subspaces of the spaces ${\mathbb{C},\mathbb{C}}$ composing $F_4$ but we also need to check that these subspaces are preserved by the maps and that the map $0 \rightarrow \mathbb{C}$ in $F_2$ is the restriction of the map $\mathbb{C} \rightarrow \mathbb{C}$ in $F_4$: this is obvious because the restriction of any map to $0$ is $0$. This should contrasted with the fact that $F_1$ is NOT a subrepresentation of $F_4$, despite the space that the spaces ${\mathbb{C},0}$ composing $F_1$ are naturally subspaces of the spaces ${\mathbb{C},\mathbb{C}}$ composing $F_4$. Indeed the map $\mathbb{C} \rightarrow \mathbb{C}$ in $F_4$ is non-zero and so cannot be restricted to a map $\mathbb{C} \rightarrow 0$.

About $Z$: the text explains what are the possible $Z$, how to define a notion of semistability given a $Z$ and how to obtain some $\theta_1$, $\theta_2$ from  $Z$ such that the semistability notion defined by $\theta_1$,$\theta_2$ is the same as the one defined by $Z$. Several $Z$ can define the same $\theta_1$, $\theta_2$ and so it is not possible to reverse the construction (which is obvious: locally $Z$ depends on two complex parameters whereas $\theta_1$,$\theta_2$ are two real parameters). Conclusion: the King's parameters or the central charge are two equivalent ways to define a notion of semistability for a quiver representation, the King's parameters being "more efficient" in the sense that they are less numerous.

The King's parameters are natural for a mathematician who has already seen similar slope definition in the context of vector bundles. The central charge point of view is more natural to a physicist thinking about BPS states. To know if a representation is semistable or not, only the information in $Z$ equivalent to $\theta_1$,$\theta_2$, i.e. essentially the phase of $Z$, is relevant but from a physicist perspective the full $Z$ is relevant (the module of $Z$ being the mass of the $BPS$ states for example).

answered Jul 2, 2015 by 40227 (4,660 points) [ revision history ]
edited Jul 2, 2015 by 40227

@40227 What about $F_0$? I cannot use the definition of the slope for this representation since the denominator is zero. What is the idea there?

Also, do you know what are the functions of the maps $Z(\vec{\delta})$ and $Z(\vec{d})$? What are they rules/formulas because it is not given in the text I quoted, or they are to be found?

P.S. It is said in the notes and you also say that $F_2 \subset F_4$. Indeed ${0, \mathbb{C}} \subset \{ \mathbb{C}, \mathbb{C} \}$. Now, how is the restriction of $z$ to $0 \subset \mathbb{C}$ giving the sub representation?

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