User talk:JackSchmidt/Archives/2009/03

Special Unitary group over a finite field
Hi! You responded to my question over at Reference_desk/Mathematics, and I was hoping you wouldn't mind clear something up. You said α(A) is the matrix whose (i,j)th entry is the qth power of the (j,i)th entry of A. However, I fail to see how this means that α(A) is the inverse of A. Direct calculation of Aα(A) with some general matrix A just gives a mess of formulae with no obvious way of simplifying to the identity. Am I missing something really obvious? SetaLyas (talk) 21:29, 26 February 2009 (UTC)
 * Yes. The unitary group is defined to consist of those matrices for which α(A) is the inverse of A. It's not true in general. It's just the same as with the standard unitary group over the complex numbers, only there the map α is the conjugate transpose. Algebraist 22:08, 26 February 2009 (UTC)
 * For instance, when A is the identity matrix, then α(A) is the inverse of A, so the identity matrix is contained in the unitary group. However, when A is upper triangular (and not diagonal) like [1,1;0,1], then α(A) is lower triangular, so not the inverse of A, and is not contained in the unitary group.  If you happen to be looking at a reference (such as Groves's book, GAP, magma, or lots of the finite group world) where an upper triangular matrix is contained in the unitary group, let me know and I can explain that (they use a different definition that gives isomorphic groups). JackSchmidt (talk) 22:22, 26 February 2009 (UTC)
 * No I'm not really looking at any particular reference, I'm just trying to work out the structure of PSU for a project I'm doing. I also need to find things like the order, centre, commutator, structure of Sylow 2-subgroups etc. Could you set me on the right track, or maybe direct me to a useful book? SetaLyas (talk) 23:46, 26 February 2009 (UTC)


 * For the "classical groups", PSL, PSU, PSp, PSΩ, I like this book as a reference:
 * The problem being that it talks about geometry for a long time, and will definitely talk about Hermitian forms when defining the unitary group. You only need to know: the identity matrix/standard inner product is a Hermitian form, and all Hermitian forms over a finite field are equivalent to it (or maybe to a scalar multiple of it), but in any case, will define an isomorphic group.
 * The problem being that it talks about geometry for a long time, and will definitely talk about Hermitian forms when defining the unitary group. You only need to know: the identity matrix/standard inner product is a Hermitian form, and all Hermitian forms over a finite field are equivalent to it (or maybe to a scalar multiple of it), but in any case, will define an isomorphic group.


 * I include below a lot of details about the group. Feel free to ignore them and read through the books.


 * The center of PSU is always the identity (general fact: If G is a perfect group, then the center of G/Z(G) is the identity; special fact SU is perfect except for SU(2,2) which is S3 and has identity center anyways), the center of SU(n,q) is cyclic (general fact: an irreducible subgroup of GL has a locally cyclic center) of order gcd(n,q+1). The commutator subgroup of PSU is PSU, unless you ask about PSU(2,2), where it has index 2.  The commutator subgroup of U is SU (I think over a general field), and the commutator subgroup of SU is SU (except for SU(2,2) = PSU(2,2)).  The center of U consists of scalar matrices z*I such that zα(z)=1, and the map z → zα(z) is a homomorphism from the group of units of the "complex" field of size q2 to the group of units of the "real" field of size q called the norm, and its kernel has size q+1.  The center of SU consists of those central elements of U with determinant 1, hence the gcd(n,q+1).


 * The sylow 2-subgroups are described in:


 * But basically these are direct products of wreath products of dihedral groups, very similarly to Sylow 2-subgroups of the symmetric groups. In characteristic 2 there is a little change: direct products of wreath products of elementary abelian 2-groups.  The Sylow p-subgroups for p dividing the characteristic are described by Chevalley's commutator formulas (which I don't know of a PSU specific reference; Carter's Simple Groups of Lie Type is very good if you are learning all of the finite simple groups of Lie type; I can just write it out for some small dimension).  There is a choice of Hermitian form that makes the Sylow p-subgroups for p dividing the characteristic much easier to understand (they are just upper triangular matrices with some compatibility conditions amongst their norms and real parts).


 * The Sylow p-subgroups for p neither 2 nor dividing the characteristic are described in:


 * and are again direct products of wreath products, this time of cyclic p-groups.


 * JackSchmidt (talk) 01:00, 27 February 2009 (UTC)


 * thank you so much for all that info! The example I'm specifically looking at is for the Sylow 2-subgroups of $$PSU(3,2^n)$$, so sadly the JoA link doesn't really work. The links it gives at the end for tackling the $$q=2^n$$ case don't seem to be available on the net, and aren't in my uni library. I'm really starting to get a bit stuck here. I only know the order of the Sylow 2-subgroups, and don't know how to work out anything else (I'm looking to find the generators, exponent, commutator, centre). I don't even know the general form of its elements! Is there anything else you could think of that could help me? SetaLyas (talk) 15:05, 2 March 2009 (UTC)


 * For three dimensions you are in luck! Page 121-122 of:
 * should be extremely helpful. It parameterizes the Sylow subgroup using the same choice of Hermitian form as GAP uses, and gives the multiplication rule in a simple fashion.  Lemma 10.13 describes the commutator subgroup.
 * This Usenet sci.math thread has some information on the Sylow subgroups. "Maximal unipotent subgroup" is fancy language for Sylow p-subgroup where the characteristic of the field is p. My posts use MAGMA's Hermitian form as it eliminates some pesky minus signs.  Lucky for you the forms coincide in characteristic 2!  Unlucky for you, none of the sources uses the standard Hermitian form (the identity matrix) like you wanted to do.
 * Describing the Sylow p-subgroups of a matrix group in characteristic p is much easier if you conjugate the matrix group so that the Sylow p-subgroup is upper triangular. The books, the usenet posts, and more or less everyone will already have done this.  Here is how they do it:
 * If Φ is an invertible matrix such that α(Φ)=Φ, then Φ is called Hermitian, and there is an associated unitary group U(Φ)=U(n,Φ)=U(n,Fq,Φ)=U(n,Fq2/Fq,Φ) = { A in GL(n,Fq2) : α(A) Φ A = Φ }. Your question took Φ=1, but Taylor and GAP take Φ to be the matrix you get by rotating 90 degrees, [ 0,0,0,...,0,1 ; 0,0,0,...,1,0 ; ... ; 0,1,0,...,0,0 ; 1,0,....,0 ], and magma and I take Φ to be the same matrix but with some −1s replacing some of the 1s. More specifically, for dimension three, Φ = [0,0,1;0,1,0;1,0,0] in Taylor's book. Using such a Φ you can have upper triangular matrices in the unitary group, but when Φ=1, you cannot.  Over finite fields, all Φ are equivalent, so most people use the nicer Φ=[0,0,1;0,1,0;1,0,0].  Due to the Lie–Kolchin theorem every Sylow p-subgroup of a group of matrices over a finite field of characteristic p is triangular in some basis, so you might as well choose such a nice basis to begin with. JackSchmidt (talk) 22:12, 2 March 2009 (UTC)
 * If Φ is an invertible matrix such that α(Φ)=Φ, then Φ is called Hermitian, and there is an associated unitary group U(Φ)=U(n,Φ)=U(n,Fq,Φ)=U(n,Fq2/Fq,Φ) = { A in GL(n,Fq2) : α(A) Φ A = Φ }. Your question took Φ=1, but Taylor and GAP take Φ to be the matrix you get by rotating 90 degrees, [ 0,0,0,...,0,1 ; 0,0,0,...,1,0 ; ... ; 0,1,0,...,0,0 ; 1,0,....,0 ], and magma and I take Φ to be the same matrix but with some −1s replacing some of the 1s. More specifically, for dimension three, Φ = [0,0,1;0,1,0;1,0,0] in Taylor's book. Using such a Φ you can have upper triangular matrices in the unitary group, but when Φ=1, you cannot.  Over finite fields, all Φ are equivalent, so most people use the nicer Φ=[0,0,1;0,1,0;1,0,0].  Due to the Lie–Kolchin theorem every Sylow p-subgroup of a group of matrices over a finite field of characteristic p is triangular in some basis, so you might as well choose such a nice basis to begin with. JackSchmidt (talk) 22:12, 2 March 2009 (UTC)

Section break

 * Now I fear I'm going to look really stupid. What, exactly, is $$F_((2^n)^2)$$, and how can it be easily represented with abstract symbols? I mean, it's going to be like some binary representation $$(\alpha_1,...,\alpha_n)$$ where each αi is either 0 or 1, but that's sorta unwieldy. Also, what is the "nontrivial element of Gal(F/K)" you mention on the Usenet thread? Would I be right in saying that $$F_((2^n)^2)\cong \{a+bJ:a,b=0,1,\ldots,2^n-1\}$$ with addition and multiplication done in the obvious way, with $$\alpha(a)=a,\alpha(J)=-J=(2^n-1)J,J^2=0$$? Am I way off?
 * We don't have the book by Taylor in our libraries. So annoying! It seems it would have solved all my problems... I mean, I am starting to get what you're saying, but due to the aforementioned confusion with what the underlying field even is, I'm still getting a bit confuddled SetaLyas (talk) 22:17, 3 March 2009 (UTC)


 * I like to view finite field elements as polynomials. I even view the complex numbers as polynomials:  If we take the real field for granted, then every complex number is just a polynomial in some weird number "i", a + b*i + c*i^2 + d*i^3 + ..., but where we set i^2 = -1.  This defines a fairly nice structure in general.  All the elements of the field are polynomials, but we "reduce" the polynomials mod a "defining polynomial".  For the complexes, the defining polynomial was x^2 = -1.
 * For the field with four elements, we take the field with two elements for granted (elements are 0 and 1, addition is XOR and multiplication is AND if you like bits), and there is some weird number "x", such that every other number is just a polynomial in x, a + b*x + c*x^2 + d*x^3 + ..., but where x^2 = x + 1.
 * In the complex numbers, we only need a+bi. In the field with four elements, we only need a+b*x.  In the first case a and b are real numbers, and in the second case they are from {0,1}.
 * Roughly speaking the same thing works for all finite fields, you just need to find an irreducible polynomial. For the field with 8 elements, xxx=x+1 is popular (but this can only be the "real field" since 8=2^3 and 3 is odd).  For the field with 16 elements, you can use yyyy=y+1, which can be the complex field when the real field has 4 elements, since 16=2^4 and 4=2^(4/2).  This one has the nice property that if you set x=yyyyy, then xx=x+1, so the "x" of the field with four elements is the y^5 of the field with sixteen elements.


 * The "nontrivial element of the Galois group" is easy to describe for finite fields. If the small field has q elements, then the element I mean is raising everything to the qth power, called the Frobenius automorphism since it is so cool, but seriously, multiply q copies of the field element together, that is all.
 * For instance, if we wanted to look at the field with 2 elements as the reals and the field with 4 elements as the complexes, then my "#" map looks like:
 * 0# = 0
 * 1# = 1
 * x# = x*x = x+1 (by the defining polynomial)
 * (x+1)# = (x+1)*(x+1) = xx + 2x + 1 = xx + 1 (since 2=0) = (x+1) + 1 (by the defining polynomial) = x (since 2=0)
 * If instead we really used the real numbers and the complex numbers, then (a+bi)# = a-bi is just the complex conjugate.
 * If instead we wanted to look at the field with 4 elements as the reals and the field with 16 elements as the complexes, then my "#" map is raising to the 4th power and looks like:
 * 0# = 0*0*0*0 = 0
 * 1# = 1
 * y# = yyyy = y + 1 (by the defining polynomial)
 * (y+1)# = (y+1)^4 = y^4 + 4y^3 + 6y^2 + 4y + 1 = y^4 + 1 (since 2=0) = (y+1) + 1 (by the defining polynomial) = y
 * (yy)# = (yy)(yy)(yy)(yy) = (y+1)(y+1) = y^2 + 2y + 1 = y^2 + 1
 * (yy+1)# = y^2
 * (yy+y)# = y^2+y (actually y^5 = y*y^4 = y*(y+1) = y^2+y, so here is "x")
 * (yy+y+1)# = y^2+y+1
 * (yyy)# = y^3+y^2+y+1
 * (yyy+1)# = ...
 * Hopefully you see some patterns. # behaves like complex conjugation, so that (f+g)# = f#+g# and (fg)# = (f#)(g#). JackSchmidt (talk) 22:59, 3 March 2009 (UTC)
 * Right, two things then. Firstly, the method you used for using $$x^n=x+1$$ doesn't work for any n (i.e. it fails for n=5 I think)... since I'm working over general n, is there a formula for a general irreducible polynomial that works for any n that I can use in the general calculations I'm doing? Also, I only got confused because you refered to # as the "non-trivial element of Gal(F/K)" in the usenet thing, so that made me think it is actually an element of the extension field... if it's just the Frobenius automorphism, then that's hunky-dory! 79.73.220.58 (talk) 15:35, 6 March 2009 (UTC)
 * Gal(F/K) is standardish notation for the Galois group of the extension. Algebraist 15:48, 6 March 2009 (UTC)


 * Yup, Gal(F/K) is notation for the Galois group, which is a cyclic group generated by the Frobenius automorphism. Assuming F/K is suitable for a unitary group, then [F:K]=2 and the Galois group is cyclic of order 2, have exactly two elements, the identity and the Frobenius automorphism.
 * In many contexts there are theorems that say "there is no one formula" for the defining polynomial.  I suspect this is one such context.  For the most part, you should not need a specific polynomial to do the work, if you remember that the Frobenius automorphism is a K-linear function on F and a group automorphism on the group of units of F.
 * Here are some nice defining polynomials (they have the property that if you use x for the knth polynomial, then the "x" for the kth polynomial is just x^( (p^(kn)-1)/(p^n-1) ). They also have the nice property that "x" is a generator of the multiplicative group of the field it defines.  I only give the even ones at the end since, if you use K of size 2^9 with polynomial x^9+x^4+1, then you have to use F of size 2^18, so have to know a polynomial like x^18+x^12+x^10+x+1.
 * 2: x^2+x+1
 * 3: x^3+x+1
 * 4: x^4+x+1
 * 5: x^5+x^2+1
 * 6: x^6+x^4+x^3+x+1
 * 7: x^7+x+1
 * 8: x^8+x^4+x^3+x^2+1
 * 9: x^9+x^4+1
 * 10: x^10+x^6+x^5+x^3+x^2+x+1
 * 12: x^12+x^7+x^6+x^5+x^3+x+1
 * 14: x^14+x^7+x^5+x^3+1
 * 16: x^16+x^5+x^3+x^2+1
 * 18: x^18+x^12+x^10+x+1
 * 20: x^20+x^10+x^9+x^7+x^6+x^5+x^4+x+1
 * Never end a message with a table of data. JackSchmidt (talk) 16:18, 6 March 2009 (UTC)

Thanks for the review
Thanks for looking at this for me. Could you please point me in the direction of the errors you refer to so I can fix them? Thanks. &mdash; Anonymous Dissident  Talk 20:43, 10 March 2009 (UTC)
 * I only looked casually at it; a second casual look found no more math problems. The section Quotients of composed functions might need to be deleted/reconsidered.  I tried to fix it, but the true answer is nasty.  There are three problems (only one of which is "wrong", and even then is more like a typo): you left out parentheses in "a times b - c"; you did not actually use the chain rule to expand out (f(g(x)))'; the numerator of "y" will cancel the denominator of one of the terms, making it tempting to substitute back in for y.  I tried editing that section to fix it, but could not make it look reasonable.  Your more explicit example though looked fine.  I didn't check the details on it, but that is the right form of answer.
 * If the article were in mainspace, I would probably tag it with refimprove (or improve it) since the sources given are all self-published. It probably needs one or two more categories too.  To address the "textbook" concern, make sure the article is not meant to teach the technique as such, but rather to describe how the technique is discussed in reliable sources.  If you want to create an article that teaches, then Wikipedia's sister projects Wikiversity and Wikibooks are good choices. JackSchmidt (talk) 21:08, 10 March 2009 (UTC)
 * How do you think the sourcing looks now? I've removed the examples, because I think they were primarily the reason it was textbooky. Now I just have sourced descriptions of how it is applied and how it is to be used. &mdash; Anonymous Dissident  Talk 10:29, 11 March 2009 (UTC)
 * New version looks good, including sourcing and the presentation of the derivative of a quotient of compositions. Personally, I think it is ready for mainspace.  JackSchmidt (talk) 12:10, 11 March 2009 (UTC)
 * Thanks. One final thing: how does this section look? And do you think the "general form" bit down the bottom is necessary or no (I've been having a bit of a debate in my own head about it)? Thanks again for all your help. &mdash; Anonymous Dissident  Talk 12:49, 11 March 2009 (UTC)
 * I thought it was a neat way to picture the product rule. The picture is clearer with the general form.  With the current content in the article, it looks reasonably well-balanced.  I am fine with either the numerical examples or the product rule, and the product rule section looks less likely to draw "not a textbook" criticism.  Removing both sections leaves the article too terse for my taste. JackSchmidt (talk) 17:18, 11 March 2009 (UTC)
 * Okay. I'm moving it to mainspace now, then. I think it is quite ready, and it's not as if it can't be improved further by people when it is in the content. Cheers, &mdash; Anonymous Dissident  Talk 20:50, 11 March 2009 (UTC)
 * Oh, and the title. Do you think "logarithmic differentiation" or "differentiation by taking logarithms"? I know the latter sounds less simple than the former, but I thought that it would help to make the article's topic distinct from logarithmic derivative. After all, "logarithmic differentiation" sounds like it is concerned with the differentiation of logarithms, when, really, it's about taking logarithms to make differentiation of other functions simpler. But then, "logarithmic differentiation" was used by more of my sources than "differentiation by taking logarithms". So I'm torn. &mdash; Anonymous Dissident  Talk 20:58, 11 March 2009 (UTC)
 * I would go with "logarithmic differentiation" even though the title is close to logarithmic derivative, but both choices have their irritations. JackSchmidt (talk) 21:19, 11 March 2009 (UTC)

Lagrange interpolation polynomial (mathgnome request)
I can't believe it's a redlink. Could you redirect it to Lagrange polynomial (a reasonable article) or alternatively, swap the names around? Thanks. 75.62.6.87 (talk) 09:32, 16 March 2009 (UTC)
 * This is done. Lagrange interpolating polynomial had already been made.  All I did was click on the red link and add the following text:

#REDIRECT Lagrange polynomial
 * but editing redirects can be a hassle if you are not used to it. JackSchmidt (talk) 16:11, 16 March 2009 (UTC)


 * Thanks!  75.62.6.87 (talk) 16:19, 16 March 2009 (UTC)

Move
If they are an admin, why not actually move the article? They should know that copy and paste moves are not allowed (even if they are the only contributor). There was also an objection on the move request page, so they shouldn't have done it anyways.  TJ   Spyke   03:12, 23 March 2009 (UTC)
 * I wasn't objecting, it was some IP who did it. I just expressed my concern about whether the proposed title violates MOS or not. Since the IP objected, the move request has needs to go through the full move request process.  TJ   Spyke   03:22, 23 March 2009 (UTC)

A little pressie...
Link to the scanned paper I was talking about ^_^ SetaLyas (talk) 13:35, 29 March 2009 (UTC)