I am amateur - mathematics is my hobby, and I find some strange structure working with toy matrices structure so I try to ask some questions regarding it. Let me allow to introduce some structure which I do not understand.

Suppose we have finitely presented monoid with unity $M$ with two generators say $g_1,g_2$. Lets relations for this monoid would be $Rel = \{g_1^2 = e , g_2^2 = g_2 \}$ where $e$ is unit element of monoid. So we have monoid $M$ to be quotient of free monoid $F$ by relations $M = F/ Rel$. $M$ is infinite. Words in $M$ has structure "$stststststst...$" etc. Rather boring ;-)

Now I want to define ring $G$ over such monoid. Lets play with field of real numbers R, as a background field. So we have ring $G = R[M]$. Suppose we are able to find such element, let's call it $g_3$ in R[M] that the following equations are satisfied:

(1) $g_i g_j = c_{ijk}g_k$ where $i,j,k=1,2,3$ just like in Lie algebra structure.

Note that $g_3$ is not element of monoid $M$ but is element of ring $G$. Also there is no antisymmetry relations for $c_{ijk}$. Then note, that from (1) we have that every element in $R[M]$ is linear combination of set of "generators" $\{ e,g_i \} , i=1,2,3$.

In one sentence within ring $R[M]$ we have some structure which allows us to easily compute every polynomial formula as it after some evaluations may be always turned into linear combination of generators. But such generators of $R[M]$ are different that generators of base monoid $M$.

Do You know any references where I may find examples of such structures? How they are called? They are examples of what? Are there any computer algebra systems which compute with such structures?

Additional remarks:

@ Darji - "just like Lie algebra" is about formal structure. It reminds me definition of Lie algebra, but of course $c_{ijk}$ is not antisymmetric nor Jacobi identity is satisfied so of course it is not Lie algebra.

@ Darij - Of course in general there is no associativity. In case I am interesting in this structure is associative, as it follow from simple algebra monoid which is associative, and by R[M] I mean formal combinations $\sum a_i g_i$ and combinations of its multiplies as in section "two simple examples" in http://en.wikipedia.org/wiki/Group_ring So we have noncommutative ring over monoid which is associative, has unity, and $c_{ijk}$ in j,k has both symmetric and antisymmetric components.

Further clarifications: Structure I tried to describe consists of multiplicative monoid, and ring over it in reals. In this ring every polynomial has linear decomposition in "basis" $g_i$, somehow as in vector space. In ring every ring element allows such decompositions ( but not every linear combination of $g_i$ is ring element so it is not linear space). What is that? Do You know examples of such structures?

@Scott: You are right I am very bad English writer. Thank You for being so polite. So I will wrote it in the most explicit way I can.

I have finitely presented noncommutative monoid with unity and two generators $g_1,g_2$: M = F/Rel where $Rel = \{g_1^2 = e , g_2^2 = g_2 \}$, $e$ is unit element and $F$ is free monoid over two generators. Because of relations $Rel$ every elemet in monoid has form for example $g = g_1g_2g_1g_2...g_1g_2$ ( alternating finite sequence with subscripts 1212... or 2121...). Different monoid elements contains different number of multiplications. It is very simple although infinite multiplicative structure.

Then I consider monoid ring over reals $R[M]$. Every element in $R[M]$ has form:

(1) $t = r_1g_1 + r_2g_2 r_3g_1g_2+r_4g_2g_1+ r_5g_1g_2g_1 + ...+r_p g_ig_kg_i...g_s+ ...$ and so on. $r_i \in R$ and $g_i \in M$.

Note that in general monomial element $g_ig_k...g_s$ every subscript has value in $\{1,2\}$ and no two following each other subscripts are the same ( they alternate like in sequence like $1212..$ or $2121..$. Of course this is standard ring definition.

In structure, I would like to describe You here, I have strange additional property: there is element $g_3$ in ring $R[M]$ ( but it is not monoid element!) which allows following decomposition:

For every $r \in R[M]$ we have

$r= r_0 e + r_1 g_1 +r_2 g_2 +r_3 g_3$

Look: there are only four terms in decomposition, even if You decompose general ring element in the form of (1). However after such decomposition I may only multiply such elements and not add them. So in fact decomposition as above, I trying to treat as some kind of "parametrization" of ring elements. Is this interesting?

As far as I know this is not standard ring property - maybe I am wrong. If I think about for example polynomial ring (that in simple case is real ring over multiplicative monoid generated by one generator $x$) such decomposition is not possible.

So I ask You if that structure was described in literature? Is it special kind of some known structure? Where to find something about it?

Thank You all for Your remarks!

  • $\begingroup$ What do you mean by "just like in Lie algebra structure"? $\endgroup$ Feb 5 '10 at 11:01
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    $\begingroup$ Besides, I would hesitate to call R[M] a ring unless it satisfies associativity, which is a constraint on $c_{ijk}$. Do you impose this constraint? $\endgroup$ Feb 5 '10 at 11:01
  • 3
    $\begingroup$ Your language is very difficult to understand, and I can't tell what you want. Every element of a monoid ring is a linear combination of basis vectors, for any basis you choose. Monoid rings have a distinguished basis given by elements of a monoid, but you can choose other bases, and if you want, you can compute structure constants $c_{ij}^k$ for these new bases. Most rings are not monoid rings, and often, they do not come with distinguished bases. If you ring is not a vector space over the real numbers, the notion of real linear combination of elements may not make sense. $\endgroup$
    – S. Carnahan
    Feb 6 '10 at 23:45
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    $\begingroup$ The existence of the decomposition of elements of R[M] implies the ring is a subspace of a 4-dimensional real vector space. There are many such rings, but only finitely many monoid rings of dimension less than 5. $\endgroup$
    – S. Carnahan
    Feb 7 '10 at 21:47
  • $\begingroup$ @Scott: so as in my real case I have 4 generators ( in question I give simpler example) it probably is present in some classification, and literature: could You or someone else, give me some references? It is amazing and very interesting that there are finite number of such structures. $\endgroup$
    – kakaz
    Feb 8 '10 at 8:02

In your decomposition in terms of $g_3$, I'm assuming $r_3 \in R$, otherwise what I'm about to say is completely useless.

If you have a monoid ring that has dimension $n$, then the monoid itself must have exactly $n$ elements. What must be happening is that there is an "extra" relation implied by the other relations, but wasn't taken into account when you came up with your list of possible words. I can't think of a good example, so here's a bad one. Let's suppose you had the monoid with two generators $x$ and $y$, and the relation $x^3 = y$. You could take as your list of words any word that doesn't contain $x^3$, like $x, y, xy, x^2y, xyx, \ldots$, but then you're missing relations like $x^2 y = y x^2$.

There's a standard procedure for determining whether you are "missing" relations in that way. It's goes by the names "diamond lemma" or sometimes "noncommmutative Grobner basis". The procedure doesn't always terminate, but when it does it gives you a way of telling whether you have a basis of words for the ring.

It's easy to apply to the example you give, but in that case there are no "missing" relations, and thus the ring is infinite-dimensional over $R$, and there is no such $g_3$.

  • $\begingroup$ Thank You for Your answer. I have to think about it carefully. I have work with certain matrix monoid in fact, so I have find explicit elements $g_i$ which satisfy relations I described, but maybe I have done something wrong in algebraic translation of it. Thank You a lot for Your remark! $\endgroup$
    – kakaz
    Jan 21 '11 at 7:18
  • $\begingroup$ Feel free to add more detail, and I'll see if I can clarify. If you have a matrix monoid, then the monoid is finite, right? (I assume you mean the Rees matrix semigroup.) So your choice of representative words would then have to be finite. $\endgroup$
    – arsmath
    Jan 21 '11 at 14:11

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