Algebra generated by transformation matrices

Question

Let $T_n$ be the full transformation monoid of an $n$-set $N_n$ with elements 1,...,n consisting of all functions $f: N_n \rightarrow N_n$. We can associate to each function $f$ a matrix $M_f$ in the same way as we associate a permutation matrix to a permutation.

Question 1: Let $A_n$ be the algebra over the field of complex numbers $K$ generated by all matrices $M_f$ for $f \in T_n$. Is there an explicit description of $A_n$ up to isomorphism? What are quiver and relations for $A_n$?

It seems the dimension of the algebra $A_n$ is given by $n^2-n+1$ and the dimension of the radical of $A_n$ is $n-1$ (so $A_n$ is not semisimple).

The answer for the same question with the symmetric group is discussed here: A finite dimensional algebra associated to the symmetric group

Now let $E(M_f)$ denote the $2^n \times 2^n$ matrix that is given as the action of $M_f$, a $n \times n$ matrix, on the Grassmann algebra of $K^n$.

Question 2: Let $B_n$ be the algebra over the complex numbers generated by all matrices $E(M_f)$ for $f \in T_n$. Is there an explicit description of $B_n$ up to isomorphism? What are quiver and relations for $B_n$?

It seems the dimension of the algebra $B_n$ is given by $\binom{2 n-1}{n}$ and $B_n$ is not semisimple again. It seems that a basis of $B_n$ is given by the matrices $E(M_f)$ such that the matrix $M_f$ is totally non-negative (or equivalently $E(M_f)$ having only non-negative entries).

I was able to work out question 2 when taking the permutations instead of all transformations and in this case the answer is very nice: The resulting algebra is isomorphic to the planar hook algebra. But it was not so hard to prove because for the permutation matrices the resulting algebra is semisimple.

For $n=2$, $A_n$ and $B_n$ are both isomorphic to the ring of 2x2 upper triangular matrices over $K$, which might lead to the question whether the algebras $A_n$ and $B_n$ have finite global dimension and what it might be. $A_n$ and $B_n$ seem to be not basic for $n>2$.

It seems that $A_n$ and $B_n$ have one dimensional center, so their quivers should be connected. By the comment of Benjamin Steinberg, they are quotients of the monoid algebra of $T_n$ and thus their quivers are acyclic and the global dimension is indeed finite. For $n \leq 5$ the dimensions in this question have been verified with GAP (assuming that the rationals are a splitting field).

Answer 1

Now that I have worked this out, I will rewrite the answer cleanly. I have deleted the old answer to unclutter things.

The algebra $A_n$ is Morita equivalent to $2\times 2$ upper triangular matrices and the algebra $B_n$ is Morita equivalent to the square-zero algebra with quiver the Dynkin quiver consisting of a straightline path with $n$ vertices. Moreover, the semisimple quotients of $A_n$ and $B_n$ are the corresponding subalgebras generated by the corresponding representations of $S_n$, asked by the OP in the previous question (linked in the OP).

All of this can be deduced from my paper The Global Dimension of the full Transformation Monoid (with an Appendix by V. Mazorchuk and B. Steinberg)

Let $T_n$ be the full transformation monoid. Then there is an exact sequence of $\mathbb CT_n$-modules $$0\to \Lambda^k(\mathrm{Aug}(\mathbb C^n))\to \Lambda^k(\mathbb C^n)\to \Lambda^{k-1}(\mathrm{Aug}(\mathbb C^n))\to 0$$ where $\mathrm{Aug}(\mathbb C^n)$ is the subspace of vectors summing up to $0$ for $1\leq k\leq n$. The modules $\Lambda^k(\mathrm{Aug}(\mathbb C^n)$ are simple $\mathbb CS_n$-modules and hence simple $\mathbb CT_n$-modules. The middle module $\Lambda^k(\mathbb C_n)$ is a projective indecomposable $\mathbb CT_n$ module (this is proved in my paper, where I write down the primitive idempotent).

For $A_n$ observe it consists of the matrices whose columns all have the same sum, which is $n^2-n+1$-dimensional. The radical consists of those matrices whose rows and columns sum to $0$, which has dimension $n-1$. It has two simple modules, $\mathbb C$ and $\mathrm{Aug}(\mathbb C^n)$ of dimensions $1$ and $n-1$ and there is a nonsplit exact sequence as above. Dimension considerations show the quiver has two vertices and one edge. Hence $A_n$ is Morita equivalent to $2\times 2$ upper triangular matrices.

To explain the $B_n$ case, we can write $1=e+f$ with $ef=fe=0$ and where $e$ is a sum of orthogonal primitive idempotents giving projective indecomposable modules isomorphic to $\Lambda^k(\mathbb C^n)$ for some $0\leq k\leq n$ and $f$ is a sum of orthogonal primitive idempotents giving projective indecomposable modules not isomorphic to any $\Lambda^k(\mathbb C^n)$. I claim that $B_n\cong e\mathbb CT_ne=\mathbb CT_ne$ and is a hence projective left $\mathbb CT_n$-module. Thus we can compute $\mathrm{Ext}$ between $B_n$-modules exactly as we do in $\mathbb CT_n$.

Indeed, the above exact sequences show that the projective indecomposable modules $\Lambda^k(\mathbb C^n)$ only have composition factors of the form $\Lambda^j(\mathrm{Aug}(\mathbb C^n))$ with $j\geq k$. Thus $(1-e)\mathbb CT_ne=f\mathbb CT_ne=0$. But then it follows that $\mathbb CT_ne= (1-e)\mathbb CT_ne\oplus e\mathbb CT_ne=e\mathbb CT_ne$, and so we have a homomorphism $\mathbb CT_n\to e\mathbb CT_ne$ given by $a\mapsto ae$ with kernel $\mathbb CT_nf$ and note that $\mathbb CT_nf$ is the annihilator of $\mathbb CT_ne=e\mathbb CT_ne$. Since $\Lambda(\mathbb C^n)=\bigoplus_{k=0}^n\Lambda^k(\mathbb C^n)$ is a projective module with the same indecomposable summands as $\mathbb CT_ne$ (but with different multiplicities), it follows that $\Lambda(\mathbb C^n)$ also has annihilator ideal $\mathbb CT_nf$ and so $B_n\cong e\mathbb CT_ne$.

In my paper, I compute $\mathrm{Ext}$ between all the simple $\mathbb CT_n$-modules of the form $\Lambda^k(\mathrm{Aug}(\mathbb C^n))$ and basically $\mathrm{Ext}^q_{\mathbb CT^n}(\Lambda^k(\mathrm{Aug}(\mathbb C^n),\Lambda^j(\mathrm{Aug}(\mathbb C^n))$ is $0$ unless $q=j-k$, in which case it is $1$.

From the above Ext computations, it follows the quiver of $B_n$ is a straightline with $n$-vertices and these are radical-square zero algebras.

Also, since the irreducible representations of $A_n$ and $B_n$ restrict to irreducible representations of $S_n$ it is straightforward that the semisimple quotients are the subalgebras generated by the image of $S_n$ in each algebra (note the above exact sequences split over $S_n$).

For $B_n$, the simple modules $\Lambda_k(\mathrm{Aug}(\mathbb C^n)$ have dimensions $\binom{n-1}{k}$ and hence the semisimple quotient has dimension $\binom{2n-2}{n-1}$.

From the description of the quiver, the dimensions of the simples and the fact that it is radical squared zero, I think you get that the dimension is $\binom{2n-2}{n-1}+\sum_{i=1}^{n-1}\binom{n-1}{i-1}\binom{n-1}{i}=\binom{2n-1}{n-1}$ using the Vandermonde identity.

Added. In fact, I have an unpublished paper (in progress but stalled due to me getting too busy) with Stuart Margolis and Itamar Stein on this algebra and in any event its quiver with relations is the Dold-Kan theorem. Let $O_n$ be the submonoid of $T_n$ consisting of order-preserving maps. This is a monoid of size $\binom{2n-1}{n-1}$ and its algebra is precisely $B_n$. I had not realized before it is a quotient algebra of $\mathbb CT_n$ and also a corner. The basis described in the OP is the one corresponding to the images of the elements of $O_n$ under this representation since order preserving maps will never turn a wedge negative, while if a map fails to preserve the order of $i<j$, then it will negate $e_i\wedge e_j$.

In our unpublished paper, we observe that Dold-Kan says that the representations of this monoid are the same as chain complexes with $n-1$ maps and $n$ vector spaces (over any field). We also have a monoid proof of Dold-Kan (which is closely related to a proof of Ross Street, done independently). Anyway, we observe that $KO_n$, for any commutative ring $K$, is the endomorphism ring of the chain complex of the $n-1$-simplex over $K$, which over $\mathbb C$ is, of course, $\Lambda(\mathbb C^n)$ equipped with its natural differential. So now I see that the image of $\mathbb CT_n$ under the natural action of $T_n$ by simplicial maps is the same as the image of $\mathbb CO_n$ in characteristic $0$, and hence $B_n\cong \mathbb CO_n$. I'm not sure why it took me so long to remember my own work.

So in summary $B_n$ is isomorphic to the monoid algebra $\mathbb CO_n$ and can also be identified with all $\mathbb C$-linear maps on $\Lambda(\mathbb C^n)$ that preserve the grading (i.e., leave each $\Lambda^k(\mathbb C^n)$ invariant) and commute with the differential $$d(e_{i_0}\wedge\cdots \wedge e_{i_k}) = \sum_{j=0}^k(-1)^j(e_{i_0}\wedge\cdots\wedge \widehat {e_{i_j}}\wedge\cdots \wedge e_{i_k})$$ (where hat means omit). So that is the concrete description.

Also this description is valid if you replace $\mathbb C$ by any commutative ring with unit $K$, but now one really needs to use the $O_n$-theory because the $T_n$-theory can break down over nonfields or fields of bad charactersitic.