\((\alpha_{I},\beta^{J})\) is the sympletic basis of the cohomology \(H^{3}\), so \(\int \alpha_{I} \wedge \beta^{J} = \delta_{I}^{J}\). \(\eta^{a}\) is the basis of \(H^{1}\) and so each of the sympletic basis elements can be written as a linear combination of terms of the form \(\eta^{abc} = \eta^{a} \wedge \eta^{b} \wedge \eta^{c}\), such as (for the particular case I'm looking at) \(\alpha_{1} = \eta^{235}\) and so \(\beta^{1} = \eta^{146}\). \(\iota_{a}\) is the set of interior products, such that \(\iota_{a}(\eta^{b}) = \delta^{a}_{b}\) and just as the symplectic 3-form basis is built from cubic combinatiosn of \(\eta^{a}\) we can build a (somewhat) dual set of interior products, \(\iota_{\alpha_{I}}\) and \(\iota_{\beta^{J}}\) by the definition/requirement \(\iota_{\alpha_{I}}(\alpha_{I}) = 1\) and likewise for the \(\beta\). Again, in my specific case I'd have \(\iota_{\alpha_{1}} = \iota_{532}\) and \(\iota_{\beta^{1}} = \iota_{641}\). The issue I'm wondering is what do I get when I apply these interior products to the other terms, such as \(\iota_{\beta^{J}}(\beta^{K})\), \(\iota_{\alpha_{J}}(\beta^{K})\), \(\iota_{\beta^{J}}(\alpha_{K})\) and \(\iota_{\alpha_{J}}(\alpha_{K})\). I'm pretty sure that given the defining property \(\int \alpha_{I} \wedge \beta^{J} = \delta_{I}^{J}\) then I'd have the following but I'm struggling to see a way to prove it and obviously confirming it for a particular case isn't good enough : \(\iota_{\beta^{J}}(\beta^{K}) = \delta_{J}^{K} = \iota_{\alpha_{J}}(\alpha_{K})\) , \(\iota_{\alpha_{J}}(\beta^{K}) = 0 = \iota_{\beta^{J}}(\alpha_{K})\)
Turns out I'm right in my guess and I can prove it by writing the 3-forms and the interior forms in terms of components. \(\alpha_{I} = (\alpha_{I})_{abc}\eta^{abc}\), \(\beta^{J} = (\beta^{J})_{ijk}\eta^{ijk}\), \(\iota_{\alpha_{I}} = (A_{I})^{abc}\iota_{cba}\), \(\iota_{\beta^{J}} = (B^{J})^{ijk}\iota_{kji}\) Using \(\int \alpha_{I} \wedge \beta^{J} = \delta^{J}_{I}\) you get a relatonship between the \( (\alpha_{I})_{abc}\) and \( (\beta^{J})_{ijk}\). Similarly for \(\iota_{\alpha_{J}}\alpha_{I}\) between the \( (\alpha_{I})_{abc}\) and \((A_{I})^{abc}\) and likewise for the \(\beta\) and then you find that \(\iota_{\beta^{J}}\alpha_{I}\) is expressed in terms of known coefficients which can be recombined into \(\int \alpha_{I} \wedge \alpha_{J}\), so \(\iota_{\beta^{J}}\) is equivalent to \(\int_{B_{J}}\) where \(B_{J}\) is the 3-cycle associated with the \(\alpha_{J}\). All fits together nicely Please Register or Log in to view the hidden image! DRZion, just because the big boys are using words you don't understand doesn't mean you have to make an arse of yourself in a desperate attempt for attention. Now run along and go play in the sandbox. And try not to eat any of it.
I'm bumping this because it's all the same kind of stuff and I might as well keep it in a single thread. I,J range over 0 to N. By definition \((\alpha_{I},\beta^{J})\) is a sympletic basis and obeys \(\int \alpha_{I} \wedge \beta^{J} = -\int \beta^{J} \wedge \alpha_{I} = \delta^{J}_{I}\). As such there's a matrix g associated to this, with entries \(g_{IJ} = g(\alpha_{I},\alpha_{J}) = \int \alpha_{I} \wedge \alpha_{J} = 0\) and \(g_{I}^{\phantom{I}J} = g(\alpha_{I},\beta^{J}) = \int \alpha_{I} \wedge \beta^{J} = \delta_{I}^{J}\) etc, so \(g = \left( \begin{array}{cc} 0 & 1 \\ -1 & 0 \end{array} \right) \, \otimes \, \mathbb{I}_{N+1}\) where \(\mathbb{I}_{N+1}\) is the N+1 by N+1 identity matrix. If I apply a linear transformation to the basis \((\alpha_{I},\beta^{J}) \to (\tilde{\alpha}_{I},\tilde{\beta}^{J}) = M\cdo(\alpha_{I},\beta^{J})\) then g is invariant if M is a sympletic matrix, ie \(\tilde{g} = M^{T}\cdot g \cdot M = g\). Now my question is if I'm given two different sets of 3-form bases \((\alpha_{I},\beta^{J})\) and \((\tilde{\alpha}_{I},\tilde{\beta}^{J})\), which have \(\tilde{g} = g\) can I be certain that there's an M such that \( (\tilde{\alpha}_{I},\tilde{\beta}^{J}) = M\cdo(\alpha_{I},\beta^{J})\)? If not, how would I go about finding the other cases or working out how many there are, is how many equivalence classes are there for the \(\sim\) defined by "\( (\tilde{\alpha}_{I},\tilde{\beta}^{J}) \sim (\alpha_{I},\beta^{J})\) if there exists an M where \( (\tilde{\alpha}_{I},\tilde{\beta}^{J}) = M\cdo(\alpha_{I},\beta^{J})\)". Comments or book suggestions would be greatly appreciated unless you're a crank, in which case go away. Posts have already been deleted from this thread because some cranks can't help but expose their dislike for people who actually read books.
Alphanumeric: does this have a relation to symmetrization of group elements and cosets? Your linear transformation resembles an \( Hg_1.s = Hg_2 \), that looks like a Schreier coset relation for the edge set \( (Hg_1,Hg_2) \). Please shoot this down if it's completely off track.
I would guess not since I have no idea what you're referring to. I have a vector \(v = \left(\begin{array}{c} \alpha_{0} \\ \vdots \\ \alpha_{N} \\ \beta_{0} \\ \vdots \\ \beta_{N}\end{array}\right)\). I can define a bilinear form g by \(g_{ij} = \int v_{i} \wedge v_{j}\). I can do a linear transformation of v by \(v \to v' = L\cdot v\) so \(v_{j} \to v'_{j} = L^{i}_{j}v_{i}\). This gives me a new g, g' such that \(g'_{ij} = \int v'_{i} \wedge v'_{j}\). I've defined the alphas and betas so they are sympletic, as given in previous posts and so if \(g' = L^{\top} \cdot g \cdot L = g\) then the entries of v' define a sympletic basis and L is a sympletic matrix. Conversely, if L is sympletric then g' = g by construction. BUT if I have a g' and a g, defined from v' and v respectively, such that g' = g can I say there is an L such that v' = L.v ?
Are you actually looking for a solution to a combinatorial problem, and does that mean something like GAP? Have you heard of Jane Gilman or the Gilman-Patterson theorem? Your Id matrix resembles what Gilman calls an intersection matrix of a corresponding "action matrix" M of an h adapted basis on the adapted canonical form (I don't know what that last is exactly). They use Schreir-Reidemeister "rewriting" in something I happen to have encountered, incidental to some papers on the Schreir-Sims algorithm.
I'm looking for a theorem or whatnot which states how many inequivalent (by which I mean not related by a sympletic transformation) sympletic bases there are for \(H^{2k+1}\) on a manifold of dimension 4k+2 (in this case k=1). If any v and v', as previously defined, are related by a sympletic matrix L, ie v' = L.v, then there's only one equivalence class for the relation v~v' if v' = L.v, with v having the g previously defined. I would imagine that it might be possible to attack this in a brute force manner for a particular manifold and cohomology but I'm looking for something more general. Problems like this often relate to the topology of the manifold in some way (after all, the dimension of the cohomologies relate to Betti and Hodge numbers) and perhaps there's some theorem like "If the Betti numbers of a manifold satisfy [something] then there are [something] equivalence classes for ~". It's somewhat academic, in the sense I do not need to know the answer to do what I am currently doing but its something which might be worth looking at in the context of my work should the answer be there's more than 1 equivalence class for particular manifolds. The name Gilman rings a bell but nothing concrete. Having checked her webpage I think I might be thinking of someone else, her area of research is quite distant from anything I'm doing or have sat lectures on.
The more I look at this paper by Gilman, the more I think it's definitely connected to your question, and the method or gadget you're after as well: http://arXiv.org/abs/math/0701286v1
