Algebras with a self-dual congruence lattice The well known (Mal'tsev) conditions that characterize certain properties of the congruence lattice of an algebra. The existence of a 3-ary term $p$ together with familiar identities $p(x,y,y) \sim x \sim p(y,y,x)$, is equivalent to congruence permutability, for example. And imilar conditions exist for other lattice theoretic properties, such as distributivity and modularity.
Is there a term-condition, or an identity, that characterizes (or implies) varieties where every algebra has a self-dual congruence lattice? 
A second question is if there is a Mal'tsev condition that implies that the every algebra has a complemented congruence lattice.
For example, the congruence lattice of a finite abelian group is self-dual. Or the congruence lattice of a finite dimensional vector space.
 A: This question is about the following two properties of varieties:


*

*The property that all members of $\mathcal V$
have self-dual congruence
lattices.

*The property that all members of $\mathcal V$
have complemented congruence
lattices.
As Pedro observed, if all members have complemented congruence lattices,
then the subdirectly irreducible members must be simple [in a nonsimple
SI the monolith has no complement]. We can also say something about the SI's
under the assumption that all members have self-dual congruence lattices.
First note that if all members have self-dual congruence
lattices, then any interval in a congruence lattice is also self-dual. This follows from repeated use of self-duality
together with the fact that upper intervals are congruence lattices of
quotients.
Now assume that $\mathcal V$ has the property that its members have
self-dual congruence lattices. The bottom element of the congruence lattice
of an SI ${\mathbf S}$
is completely meet irreducible, so the top element must be completely join
irreducible. By the observation of the last paragraph, every nontrivial
upper interval is self dual. Since these intervals $[\sigma,1]$ have completely
join irreducible top element, they must also have completely
meet irreducible bottom element. Consequently every non-top
element in the congruence lattice of ${\mathbf S}$ is completely
meet irreducible, and similarly every non-bottom
element is completely join irreducible. The only algebraic lattices where
every non-top element is completely meet irreducible and every non-bottom
element is completely join irreducible are finite chains.
Conclusion: If every algebra in
${\mathcal V}$
has a self-dual congruence lattice, then the congruence lattice
of any SI in $\mathcal V$ is a finite chain.
Continuing to assume that members of $\mathcal V$ have self-dual congruence lattices, if ${\mathbf A}\in {\mathcal V}$ is nontrivial and $\alpha$
is a proper congruence on $\mathbf A$, then $\alpha$ can be enlarged
to $\beta$ so that ${\mathbf A}/\beta$ is SI. By the result of
the last paragraph, we may enlarge $\beta$ further to $\gamma$
if necessary where ${\mathbf A}/\gamma$ is simple. This shows that
every proper congruence $\alpha$ on a nontrivial algebra in $\mathcal V$
can be enlarged to a coatom $\gamma$. By self-duality we get that every nonzero congruence must lie above an atom.
This is a strong property to impose throughout a variety. It
is investigated in this paper:
Atomicity and nilpotence, Canad. J. Math. 42(1990), 365-382.
A result found there is that if every nonzero congruence on every
nontrivial algebra in $\mathcal V$ lies above an atom, then the ascending
central series for any ${\mathbf A}\in {\mathcal V}$ reaches the top congruence,
although perhaps after transfinitely many steps.
What it means here is that if we are in the case where $\mathcal V$ has
self-dual congruence lattices, then SI's in
$\mathcal V$ have congruence
lattices that are finite chains, and the SI algebras themselves are nilpotent.
(It is plausible that the SI's must even be abelian, 
but I see how to prove this only
for congruence modular varieties at the moment.)
If we are in the case where $\mathcal V$ has
complemented congruence lattices, then SI's in $\mathcal V$ are
simple and, as one can show by following the argument in the above paper, the SI's are abelian. In this case $\mathcal V$ is an abelian variety. 
Now for the answers these questions:
Is there is a Maltsev condition that implies that the
every algebra has a complemented congruence lattice?
and
Is there is a Maltsev condition that implies that the
every algebra has a self-dual congruence lattice?
Claim: The only such Maltsev conditions are those
expressing that the variety is trivial.
This Claim rests on the following
Observation: If $\Sigma$ is a Maltsev condition
that is satisfied by a nontrivial variety, then it is satisfied
by a nontrivial discriminator variety. [If $\mathcal V$ satisfies
$\Sigma$, just add the discriminator $d$ to each member of $\mathcal V$
and generate a variety ${\mathcal V}^d$ with the resulting algebras.
${\mathcal V}^d$ is a discriminator variety satisfying any Maltsev
condition true in ${\mathcal V}$.]
To see how the Observation implies the Claim, assume
that $\Sigma$ is a Maltsev condition implying either that
congruence lattices are self-dual or implying that congruence
lattices are complemented. Let $\mathcal V$ satisfy $\Sigma$.
The discriminator variety ${\mathcal V}^d$ satisfies $\Sigma$, so
its SI's in ${\mathcal V}^d$
are nilpotent. But the only nilpotent members of a discriminator
variety are trivial, so ${\mathcal V}^d$ is trivial. $\mathcal V$
must also be trivial.
A: Some thoughts around the problem concerning complements.
$\newcommand{\calV}{\mathcal{V}}$
Let $\calV$ be a variety. The condition (call it C) that every algebra in $\calV$ has a complemented congruence lattice is rather strong. For instance, it implies semisimplicity: Every subdirectly irreducible algebra $A$ must be simple. This happens since every such $A$ has a least nonzero congruence, which must be the greatest one in order to have a complement.
But more interestingly, this property is “incompatible” with congruence-distributivity; this means that the only congruence-distributive variety satisfying C is the trivial one. One proof of this fact goes along the following lines.
Assume $\calV$ is a congruence-distributive variety satisfying C. Let $A$ be a simple member of $\calV$, and $B$ the Boolean algebra of finite and cofinite subsets of $\mathbb{N}$. You can construct the Boolean power (check the online book A Course in Universal Algebra for this) $P=A[B]^*$, and in the case at hand the lattice of congruences of $P$ is isomorphic to that of $B$ (see, for instance, A. G. Pinus, Boolean Constructions in Universal Algebras, Lemma 3.2 and Corollary 3.1). But it is easy to see that the lattice of congruences of $B$ is not complemented; actually, the congruence associated to the cofinite filter has no complement.
On the other hand, if you have a congruence-permutable variety satisfying C, every congruence of any of its members must be a factor congruence. This is also rather suspicious.
