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 " tommy quaternion " tommy1729 Ultimate Fellow Posts: 1,493 Threads: 356 Joined: Feb 2009 01/18/2021, 11:03 PM (This post was last modified: 01/21/2021, 11:40 PM by tommy1729.) I always like stuff like bicomplex numbers , hypercomplex numbers etc And some of you here do to. So I thought I share my " tommy quaternion " with you. Do not confuse with the normal quaternion or hyperbolic quaternion. The nonreal units are $A,B,C$ And the rules for the units are : $A*B=B*A=C,A*C=C*A=-B$ $B*C=C*B=A$ $A*A=B*B=-1$ $C*C=1$ The distributive property holds. And clearly this is a commutative number system. But not associative. A pair of zero divisors are $(A+B)(A-B)=A^2 - B^2 = -1 - -1 = 0$ The word " anti-associative " comes to mind. Im considering doing tetration with them. Feel free to comment. Regards tommy1729 Tom Marcel Raes marraco Fellow Posts: 100 Threads: 12 Joined: Apr 2011 01/21/2021, 12:49 PM Can you post the matrix representation? I have the result, but I do not yet know how to get it. tommy1729 Ultimate Fellow Posts: 1,493 Threads: 356 Joined: Feb 2009 01/21/2021, 01:17 PM (This post was last modified: 01/21/2021, 11:40 PM by tommy1729.) (01/21/2021, 12:49 PM)marraco Wrote: Can you post the matrix representation? Hi marraco. These numbers have NO matrix representation because multiplication is NOT associative. So doing simple linear algebra probably will not work. This is one of the simplest algebra's that have no matrix representation. Or should I say smallest. Dimension 3 does not have this property if we require it to have some group like structure. So dimension 4 is the smallest. HOWEVER it is not the only one in dimension 4. There exists at least one other nonisomorphic one : unfortunately neither of them are closed under square root. ( $1 + 2 A + 3 B + 4 C$  has no square root in either one ) Another one is  $A*B=B*A=C,A*C=C*A=B$ $B*C=C*B=A$ $A*A=B*B=-1$ $C*C=1$ Thank you for your interest. regards tommy1729 MphLee Fellow Posts: 184 Threads: 19 Joined: May 2013 01/21/2021, 09:27 PM (This post was last modified: 01/21/2021, 10:59 PM by MphLee.) Hi, why you don't specify that $C*B=A$ in your list of identities? You forgot to add it or it is possible to derive it from the others? Also, you see it as a non-associative algebra over the reals or the complex? Trivial question (sorry but I'm new to linear algebra): being an algebra means that $\mathbb T$ is a $k$-vector space of dimension d=4 on which is defined a bilinear application $*:{\mathbb T} \times {\mathbb T}\to {\mathbb T}$ that is not associative... but it must be bilinear because we want it distributive: An element is of the form $T=x1+yA+zB+wC$. Let  $T,R,S,U\in{\mathbb T}$, the application being distributive means that $(T+R)*(S+U)=T*S+R*S+T*U+R*U$ i.e. every translation by a Tommy quaternion is a endomorphism of  the addition group; and we also want that for every scalar $\lambda$ $\lambda R*S= R*\lambda S=\lambda (R*S)$ i.e we want it to contain a copy of the base field, i.e multiplication by scalar  is multiplication by $\lambda1$ where $1\in \mathbb T$. So the previous means that the Tommy quaternions of the form $T=x1+0A+0B+0C$ commute and "associate" with everything. But this means that the multiplication is bilinear: every element should it should have a representation.. e.g. multiplication by $T$ has a 4x4 matrix $M_T$. Obviously if it is not associative we "should not" have $M_{T*S}=M_{T}M_{S}$ in general... Probably my understanding of non-associative alg. is so poor that I'm missing something obvious. MathStackExchange account:MphLee Fundamental Law $(\sigma+1)0=\sigma (\sigma+1)$ tommy1729 Ultimate Fellow Posts: 1,493 Threads: 356 Joined: Feb 2009 01/21/2021, 11:47 PM (01/21/2021, 09:27 PM)MphLee Wrote: Hi, why you don't specify that $C*B=A$ in your list of identities? You forgot to add it or it is possible to derive it from the others? Also, you see it as a non-associative algebra over the reals or the complex? Trivial question (sorry but I'm new to linear algebra): being an algebra means that $\mathbb T$ is a $k$-vector space of dimension d=4 on which is defined a bilinear application $*:{\mathbb T} \times {\mathbb T}\to {\mathbb T}$ that is not associative... but it must be bilinear because we want it distributive: An element is of the form $T=x1+yA+zB+wC$. Let  $T,R,S,U\in{\mathbb T}$, the application being distributive means that $(T+R)*(S+U)=T*S+R*S+T*U+R*U$ i.e. every translation by a Tommy quaternion is a endomorphism of  the addition group; and we also want that for every scalar $\lambda$ $\lambda R*S= R*\lambda S=\lambda (R*S)$ i.e we want it to contain a copy of the base field, i.e multiplication by scalar  is multiplication by $\lambda1$ where $1\in \mathbb T$. So the previous means that the Tommy quaternions of the form $T=x1+0A+0B+0C$ commute and "associate" with everything. But this means that the multiplication is bilinear: every element should it should have a representation.. e.g. multiplication by $T$ has a 4x4 matrix $M_T$. Obviously if it is not associative we "should not" have $M_{T*S}=M_{T}M_{S}$ in general... Probably my understanding of non-associative alg. is so poor that I'm missing something obvious. yeah ... I forgot to add C*B sorry. And thanks for pointing it out. It is corrected now. Well matrix representation is based on representing the multiplication. But matrix multiplication is associative , however these numbers are not associative by multiplication. Functional composition is also associative btw. So easy representations might not exist. In my opinion this makes these commutative algebras interesting. regards tommy1729 MphLee Fellow Posts: 184 Threads: 19 Joined: May 2013 01/22/2021, 12:03 AM Ah I see... you didn't mean that multiplication is not a bilinear map, we can represent only the translations... you mean representable in the sense of a morphism from the Algebra multiplication to a $M_{n\times n}(k)$ or to a GL(n), e.g. like a group representation, where we study characters and so on... In fact how can we "represent" non associative operations? Sorry but I'm not very fluent with this yet.. MathStackExchange account:MphLee Fundamental Law $(\sigma+1)0=\sigma (\sigma+1)$ tommy1729 Ultimate Fellow Posts: 1,493 Threads: 356 Joined: Feb 2009 01/22/2021, 12:10 AM (This post was last modified: 01/23/2021, 01:23 PM by tommy1729.) Then again maybe dimension 4 is too small and simple to be interesting. It does not even have a square root if we use real coefficients. ( they have 8 solutions over the complex coefficients though ) Now it turns out commutative nonassociative hypercomplex numbers that have a square root must be of dimension 4n. Powerassociative is also a thing. Im not an expert in this though. Informally speaking , the main issue is constructing such cayley tables such that it does not have any " subcayleys " (the analogue of subgroup), yet do have a structure that makes the square root possible. In dimension 8 we have for instance this  (see picture) But Im not sure about the properties yet. 8 = 4*2 , so the complex might be a subgroup and then the other subgroup would be one of those 4 dim ones. ( edit : it has no subalgebra , this has been checked see follow up replies ) sqrt has not been investigated. I think it is power-associative. regards tommy1729 Attached Files Image(s)     tommy1729 Ultimate Fellow Posts: 1,493 Threads: 356 Joined: Feb 2009 01/22/2021, 12:15 AM (01/22/2021, 12:03 AM)MphLee Wrote: Ah I see... you didn't mean that multiplication is not a bilinear map, we can represent only the translations... you mean representable in the sense of a morphism from the Algebra multiplication to a $M_{n\times n}(k)$ or to a GL(n), e.g. like a group representation, where we study characters and so on... In fact how can we "represent" non associative operations? Sorry but I'm not very fluent with this yet.. Powerassociativity together with newton's binomium and tensor and nonlinear theory are useful. However Im not an expert and your questions are good. The complexity is the reason Im into this. I feel hypercomplex numbers deserve more attention. Maybe it is hidden or maybe it is not investigated sufficiently. Or both. regards tommy1729 MphLee Fellow Posts: 184 Threads: 19 Joined: May 2013 01/22/2021, 12:41 AM (This post was last modified: 01/22/2021, 12:47 AM by MphLee.) I've to look deeper on this power associativity... sounds familiar... with unitBtw I agree. Basically if we regard as hypercomplex everything that is a finite dimensional algebra, not necessarily commutative or associative, over the real, e.g. (C, H, O, S) , in the last years, doing some readings for hobby... I grow more convinced that those structures conceal a great deal of fundamentally natural properties that has to do with the very essence of the universe in some way. For example I tried, not very successfully, to follow John Baez on his blog posts about the octonions and their relevance to the standard model, or the dimensions in which some kind of hypercomplex algebras has some properties that have to do with the dimensions were spacetime appears in some obscure and esoteric string theory black magic, or still about dual numbers were we can rebuild analysis algebraically by the use of idempotent elements acting as infinitesimals. EDIT: I'm not sure that my brain can process that 8-dim caley table. I might try too look at it with more calm. I'm way less expert at it than you btw. MathStackExchange account:MphLee Fundamental Law $(\sigma+1)0=\sigma (\sigma+1)$ tommy1729 Ultimate Fellow Posts: 1,493 Threads: 356 Joined: Feb 2009 01/23/2021, 01:17 PM The 8 dimensional " tommy octonion " - or whatever we will call it - is not constructed at random nor by trial and error. It has to satisfy properties and those properties help in the construction. For instance commutative implies (x*y)^2 = x^2 * y^2. So that is " respected " in the construction. That also implies how many -1 and +1 we have in the table. Another thing is (x*x)*y = x*(x*y). This also helps alot. ( alternative algebra , jordan algebra , power associative ) I also did a subalgebra test ( the analogue of a subgroup test ). To be more specific we have dimension 8. If dimension 8 has a subalgebra then it must be one of order 4 and of order 2 ( since 2*4 = 8 analogue to subgroups ). So we only need to check for subgroups of order 2. So I checked all the A,B,C,D for being the complex imaginary i^2 = -1 and all the E,F,G for being the split complex number j^2 = 1. None of them matched so the split-complex or complex numbers are NOT a subalgebra. Hence the " tommy quaternion " is not an extension. For the subalgebra test see the 2 added pics. regards tommy1729 Attached Files Image(s) « Next Oldest | Next Newest »

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