We had to restart our SQL Server today, we had made no changes to it.

When it came back up we immediately started getting this error from the server


  DATE/TIME:    2/27/2014 3:09:31 PM
  
  DESCRIPTION:  The SQL Server performance counter 'Number of
  Deadlocks/sec' (instance 'Database') of object 'SQLServer:Locks' is
  now above the threshold of 1.00 (the current value is 2.00).
  
  COMMENT:  (None)
  
  JOB RUN:  (None)


We ran the DBCC TRACEON (1204) command and have watched the log's but it's not reporting any deadlocks.

Any idea what could trigger this to just go off? We are getting the alert every minute yet can't find any actual deadlocks.

Edit: I should add that before this reboot we had never received this error

Edit 2: We used SQL Server Profiler as well to look for deadlocks, let it run for 5 minutes over which we received 5 error alerts and when we checked the details we had NO deadlocks found.

Edit 3: March 06/2014: Ran the query and it worked, but it reports what our other details have said that we have no locks we where still getting the error above the whole time.

Thanks again for all your help!



Edit 4: March 06/2014: I ran the query and here is a sampling of the result set, I will admit I am not exactly sure what I am looking at here, that is to say I am not sure if it shows me something that I can act on or not.



Edit 5: March 07/2014: Image below shows the Alert that generates this error all of a sudden.



Thanks


Question: Are these adjoint functors to/from the category of monoids with semigroup homomorphisms?
Answer: Functor $F\circ G_{O}:\mathbf{Semigroup}^{\mathbf{1}}\rightarrow\mathbf{Monoid}$
serves as left adjoint of functor $G_{H}:\mathbf{Monoid}\rightarrow\mathbf{Semigroup}^{\mathbf{1}}$.

Does functor $G_{O}:\mathbf{Semigroup}^{\mathbf{1}}\rightarrow\mathbf{Semigroup}$
have a left adjoint? Let's say it has and let's denote it as $K:\mathbf{Semigroup}\rightarrow\mathbf{Semigroup}^{\mathbf{1}}$.
Then the composition $F\circ G_{O}\circ K:\mathbf{Semigroup}\rightarrow\mathbf{Monoid}$
should serve as left adjoint of $G=G_{O}\circ G_{H}:\mathbf{Monoid}\rightarrow\mathbf{Semigroup}$.
This according to the rule $\left(G_{O}\circ G_{H}\right)^{ad}=G_{H}^{ad}\circ G_{O}^{ad}$.
Then $F$ and $F\circ G_{O}\circ K$ must be isomorphic functors (both
left adjoints of $G$). However, if you start with an object $S\in\mathbf{Semigroup}$
having no identity and a finite underlying set, then $K\left(S\right)$
must have at least $1$ element more (the missing identity must be
'added') and $F\circ G_{O}\left(K\left(S\right)\right)=F\left(K\left(S\right)\right)$
will on its turn have $1$ element more than $K\left(S\right)$ (again
an identiy is added). Then $F\circ G_{O}\left(K\left(S\right)\right)=F\left(K\left(S\right)\right)$
has at least $2$ elements more than $S$ in contrast with $F\left(S\right)$
that has exactly $1$ element more. This shows that the functors cannot
be isomorphic. The conclusion is that $G_{O}:\mathbf{Semigroup}^{\mathbf{1}}\rightarrow\mathbf{Semigroup}$
has no left adjoint.