Finding exceptional regular representations of $\tilde{D}_4$ efficiently Let $A$ be the path algebra of the quiver $\tilde{D}_4$. I would like to find its exceptional regular representations with as little computation as possible.
Of course, we can compute the whole Auslander-Reiten quiver and compute all the $\text{Hom}_A(X,X)$ and $\text{Ext}_A^1(X,X)$ for all the regular representations, but surely there must be a better way to do it?
A conjecture I have is that an exceptional indecomposable representation must be on the border of the regular component of the AR-quiver. By the border I mean the set of vertices of the quiver such that there is only one incoming and one outgoing edge. If anyone knows a standard term for this set, please tell me in the comments! I think there are only 6 exceptional regular representations of $\tilde{D}_4$, all of which are of this form. This conjecture seems plausible to me, but I haven't managed to prove this in general.
EDIT: it seems that it's very much untrue that an exceptional module must be on the border of a regular component, but I think I managed to prove this for the components on which the Auslander-Reiten translate $\tau$ satisfies $\tau^2=\text{id}$, and with that it is indeed possible to find the exceptional modules without computing the whole quiver. I will write down an answer here as soon as I find the time.
 A: The AR quiver of the regular representations of an affine quiver consists of infinitely many "tubes". A tube of rank $r$ has $r$ modules on what you call the border. Let me number them $B_1, B_2, \dots, B_r$. Then, if I numbered them in the convenient way, there are another $r$ modules $C_1, \dots C_r$, and irreducible morphisms from $B_i$ to $C_i$ and from $C_i$ to $B_{i+1}$. Then there are another $r$ modules $D_i$ with morphisms from $C_i$ to $D_i$ and from $D_i$ to $C_{i+1}$. This pattern continues infinitely. The Auslander--Reiten translation sends the module on each level numbered $i$ to the object numbered $i-1$ (cyclically).
The AR quiver of an affine quiver consists of infinitely many rank 1 tubes and up to three tubes of higher rank. The tubes are standard, meaning that you can calculate Homs by looking at compositions of arrows in the component modulo mesh relations. $\operatorname{Ext}^1(V,V)$ is dual to $\operatorname{Hom}(V,\tau V)$, so its dimension can be calculated the same way. One discovers that the exceptional representations in a tube of rank $r$ lie in the bottom $r-1$ layers of the tube.
In $\widetilde D_n$, there are three exceptional tubes, of rank 2, 2, and $n-2$. This provides $2 + 2 + (n-2)(n-3) = n^2 -5n + 10$ exceptional representations.
A possible reference is Chapter 7 of the book "An introduction to quiver representations" by Derksen and Weyman. Though they say they don't give full details and encourage the reader to consult "Indecomposable graphs and algebras" by Dlab and Ringel, published in the Memoirs of the AMS.
Another excellent reference would be the notes by Crawley-Boevey. The answer for $\widetilde D_4$ is given on page 35.
A: I finally found the time to type up an answer. The result is much less general than I first suspected, but I found it quite surprising that the regularity of the representation and the fact that it must be exceptional end up imposing such strong restrictions on the dimension vector of the representation. Maybe someone here will find it interesting, or, at least, mildly amusing too?
Let's denote the vertices of the $\tilde{D}_4$ quiver by $1,\ldots, 5$ with a single arrow $i\to 5$ for each $i\in \{1,2,3,4\}$. The indecomposable contravariant representations $V$ are then configurations of four subspaces $V_i$ in a vector space $V_5$. Let $S_i,\ P_i$ denote the standard simple and projective representations respectively. The Auslander-Reiten translate $\tau$ induces a linear operator on the Grothendieck group $K_0$, and its matrix $T$ in the basis $[S_i], i=1,\ldots,5$ satisfies
$$T=\begin{pmatrix}
0 & 1 & 1 & 1 & -1\\
1 & 0 & 1 & 1 & -1\\
1 & 1 & 0 & 1 & -1\\
1 & 1 & 1 & 0 & -1\\
1 & 1 & 1 & 1 & -1\\
\end{pmatrix},\ T^{2n}=\begin{pmatrix}
n+1 & n & n & n & -2n\\
n & n+1 & n & n & -2n\\
n & n & n+1 & n & -2n\\
n & n & n & n+1 & -2n\\
2n & 2n & 2n & 2n & -(4n-1)\\
\end{pmatrix}$$
Let $d_i=\dim V_i$ be the dimensions of the components of a regular representation, which are also the coordinates of its class in $K_0$. Then it is neither preinjective nor preprojective, so applying $T^k$ for any $k\in \mathbb{Z}$ should give us a nonnegative vector. In particular, we find
$$ \frac{\sum_{i=1}^4 d_i}{2+\frac{1}{2n}}\leqslant d_5\leqslant \frac{\sum_{i=1}^4 d_i}{2-\frac{1}{2n}}\quad \forall n>0,$$
so $2d_5=\sum_{i=1}^4 d_i$.
Now let's use the other condition. Note that the standard resolution of $V\neq P_5=S_5$ has the form
$$0\to P_5^{r}\to\bigoplus_{i=1}^4P_i^{d_i}\to V\to0,$$
where $r=\sum_{i=1}^4d_i-d_5$, and $r$ can be safely assumed to be non-negative, since $V$ is indecomposable, because otherwise we could find some number of $S_5$ subrepresentations. Let's look at what applying $\text{Hom}(-,V)$ does:
$$0\to \text{Hom}(V,V)\to \prod_{i=1}^4 \text{Hom}(P_i^{d_i},V)\to \text{Hom}(P_5,V)^{r}\to \text{Ext}^1(V,V)\to\cdots.$$
Since $V$ is exceptional, $\text{Hom}(V,V)=k,\ \text{Ext}^1(V,V)=0.$
Now we compute $$\dim \text{Hom}(P_j,V)=d_j,$$ so by counting the dimensions we have $$1+rd_5-\sum_{i=1}^4 d_i^2=0$$
Substituting $2d_5=\sum_{i=1}^4 d_i$ into this we can get an equivalent expression
$$
  \sum_{\substack{1\leqslant i,j\leqslant 4 \\ i\neq j}}^n (d_i-d_j)^2=8.
$$
The solutions have the form $(d_1,\ldots,d_5)=(n+1,n+1,n,n,2n+1),\ n\in\mathbb{N},$ up to a permutation of the first four elements.
Let $N_{ij}$ be the indecompasable representation with $d_i=d_j=d_5=1, d_p=d_q=0$ for $\{i,j,p,q\}=\{1,2,3,4\},$ i.e. a configuration of two lines being identically mapped onto a single line. It can be manually checked that these 6 representations are exceptional and that $\tau(N_{ij})=N_{pq}$ in the previous notation.
It remains to show that all other regular representations with dimension vector of the form $(n+1,n+1,n,n,2n+1)$ (up to a permutation) can not be exceptional. Note that just by counting dimensions we find that such a representation $M$ must contain a subrepresentation $N_{12}\subset M.$ This means that $M$ and $N_{12}$ are in the same component of the Auslander-Reiten quiver. $\tau^2=\text{id}$ on this component, so by weaving the quiver we find that it consists of the representations $N_{12}=M_0,\tau(M_0)=N_{34}, M_i,\tau(M_i),\ i>0,$ with short exact sequences
$$0\to M_0\to M_i\to\tau(M_{i-1})\to 0$$
and irreducible morphisms $M_{i-1}\to M_i$ for $i>0$. Let's apply $\text{Hom}(M_i,-)$ to this sequence:
$$0\to\ldots\to\text{Ext}^1(M_i,M_i)\to\text{Ext}^1(M_i,\tau(M_{i-1}))\to 0.$$
If $M_i$ were exceptional, by the definition of the Auslander-Reiten translate we would have $$0=\text{Ext}^1(M_i,\tau(M_{i-1}))=\text{Ext}^1(M_i,\tau^{-1}(M_{i-1}))=\underline{\text{Hom}}(\tau^{-2}(M_{i-1}),M_{i})^*=\underline{\text{Hom}}(M_{i-1},M_{i})^*,$$
so we would have no arrows from $M_{i-1}$ to $M_i$ in the AR quiver, which is impossible. This argument basically shows that if an exceptional regular representation belongs to a component of the AR quiver on which $\tau^2=\text{id},$ then it must belong to its 'border'.
All in all, we have shown that the only exceptional regular representation of $\tilde{D}_4$ are the 6 representations $N_{ij}.$
