Dogtanian

Oh, I get what you did now. You took the contrapositive of the statement: instead of proving 'if K(v,w)= 0 for a ll w then v=0' you proved 'if v=/=0 then there exists a w such that K(v,w)=/=0'. I think I get it now. So to do this question fro the start you find the matrix K= 004 080 400 in the way it was done above. Then you pick a non-zero v = (a,b,c), say, and fnd a w such that K(v,w)= vKw=/=0. In the case above you found that w=(c,b,a) will work.

So, say we to show v =0 for all w.# Let w= a b c-a = (b,a,c) and v= x y z-x = (y,x,z) Then wKv= 8.a.x + 4.c.y + 4.b.z = 0 Now is it simply that for this to be true for all a,b,c then x=y=z=0 automatically?

Thanks for telling me K(X,Y) = 4tr(XY). But I thought I'd try to work it out myself anyway as I need it to show how one gets it. I am also able to work out the matrix rep of K you gave. Now, how do I go from here? Do I use this matrix and apply it to a general Y, set this to zero and show that Y must always be zero in this instance? So, I could write a general Y not in the form of a matrix like a b c-a but as a vector a b c as by doing this you clearly see by multiplying the matrix for K with such a vector and setting the answer to zero you must have a=b=c=0.

Here's my go so far at the sl_2© case....are there any mistakes? In calculating the Killing form here, I need to show that sl_2© is semisimple. So I need to show the Killing form is non-degenerate for sl_2©. Due to bilinearity, all I need to show is if K(X,Y)=0 for all X, then Y=0. I can do this by considering the basis e,f,g (1 0) (01) (00) (0-1) (00) (10) and hopefully show K(e,Y)=0, K(f,Y)=0, K(g,Y)=0 implies Y=0. Let's start by calculating adX for each basis element of each basis element. ade(e)= 0 ade(f)= 2f ade(g)= -2g adf(e)= -2f adf(f)= 0 adf(g)= -e adg(e)= 2g adg(f)= -e adg(g)= 0 This represents the matrices for ade, adf, adg as the following 3 3x3 matrices (000) (0 -20) (002) (020) (0 00) (-100) (00-2) (-1 00) (000) I can can then work out ad(e_i)ad(e_j) for all possible combinations for e_i, e_j from e,f,g. Now, I think I will get all much matrices to have zero trace(I haven't calculated then all yet), no matter whether Y is zero or not. I don't think this is what I want....have I gone wrong somewhere?

Thank you. I actually made a mistake, when I said I needed to do this for the upper triangular 2x2 matrices. I really need to do it for the derived Lie algebra of the 3x3 upper triangular matrices (and hence show the Killing from is zero for all matrices of this derived Lie algebra). But your hints have helped me out at first... Is this right: We have the basis e,f,g= 010 001 000 000 000 001 000 000 000 For a general X of the form 0 a b 0 0 c 0 0 0 we have adX(e) = -c.f adX(f) = 0 adX(g) = a.f Now, what I need to do is to show the Killing form ios zero in all cases of X, Y being matrices being made up from the basis e,f,g. Would I now be able to consider X being each of e,f,g in turn in the above and see the results for adX(e) etc in each case (since the Killing form is bilinear). From this I get ade(e)= 0 ade(f)= 0 ade(g)= f adf(e)= 0 adf(f)= 0 adf(g)= 0 adg(e)= -f adg(f)= 0 adg(g)= 0 What, if any, matrix would I be able to work out for the whoe transformation now? (I think this is where my mind is stuck) Would it be, respectively: ade, adf, adg 000 000 0-10 000 000 0 00 010 000 0 00 ? Thus we'd see that the mutliplication of any two of these matrices together would result in all zero matrices, all of whih have zero trace and all of which mean the Killing form between an two pairs of basis elements is zero. Hence by bilinearity we see that for any two matrices in the derived Lie algebra we get the Killing form to be zero, hence we see by Catan's Critrion for solvability, that the Lie algebra of all upper triangular 3x3 matices is solvable. Is this the case? I'll think about the sl_2© example now...

OK, firstly I hope this is the rigth place for my question. I'm in a bit of a problem. I need to be able to calucalte the Killing for for a Lie algebra by next wek, but I'm stuck and won't be able to get any help in 'real life' until Friday, not leaving me enough time to sort out my problem. So I was hoping someone here might be able to show me some pointers. So I have the Lie algebra of all upper triangular 2x2 matrices and am using the basis (1 0) (0 1) (0 0) (0 0),(0 0),(0 1). (I hope these matrices turn out OK: they should be 3 2x2 matrices.) The Killing form is defined as K(X,Y) = trace(adX adY) for all X,Y in the Lie algebra, where ad is the adjoint, defined on Z by adX(Z) = [X, Z] = XZ-ZX. I know I have to consider adX and adY as linear transformations and work out their matrices (and hence multiply them together and find the trace). What I'm not sure is how to go about this, and at what point do I use the bases instead of a general element from the Lie algebra? Any help would be much appreciated, as I've spent a fair bit of time on this and got no where. I also have to so the same for the special linear Lie algebra sl_2© (C=complex nubers), but I think if I have some hints for the triangular one I should hopefully be able to work the special linear one out my self.

Last I heard (a year or so back) was that the Lib Dems said they would not form any coalition Governemnt with anyone, so if there was a hung Parliament, you can expect another election within a few months (like happend some time in the 70's I believe).

Thanks for your suggestions. I will go see my lecturer about it. Only I though't I'd ask here first since it wasn't possible to go see the lecturer before the work had to be handed in (in the morning). Anyway I'd already gone on with the assumption that maximum subgroup meant the largest and it seamed to work out fine for the questions I had. Though I'm still none-the-wiser about maximal v maximum since it turned out I didn't need any results directly relating to maximal subgroups for the questions. I guess I'll find out if I was right in a couple of days...whenever I get the work back or ask the lecturer...which ever is first. Thanks again.

Doesn't anyone know? I know semigroups are quite a specialist subject, but I thought someone migth know... ...please

I knew there was something about why all this about xnot=x didn't fit...I think you got it there, yes. I'd like to see someone say that you cannot claim that 2 apples are never two apples, no matter which appels you've got. Another example. Say you have 3 chairs : one arem chir, one deckchair and one stool. Youc an say swap the stool for a dinning chair. The stool certainly is different from the dinning chair, but you still have 3 chairs.

Can anyone tell me the difference between maximal and maximum subgroups of a semigroup, if there is one at all? I seem to be a bit confused about if there is a difference at all. In my lecture notes I can only find references to maximal subgroups, whereas on a problem sheet I can only find references to maximum subgroups. I also vaguely remeber my lecturer emphasining a difference between maximum and maximal, but I was concentrating on the proof she'd just shown at the time, so I'm not sure if she was talking about a difference within semigroup theory or the difference between semigroup theroy and group theroy. Any clarification would be great

At a guess bloodhound, I say a "proper" subgroup was just a subgroup of G that in't G itself, saying that though, I don't think I've conme across the term before either...*thinks*...

At a guess bloodhound, I say a "proper" subgroup was just a subgroup of G that in't G itself, saying that though, I don't think I've conme across the term before either...*thinks*...

Maybe we are thinking of all shadows as being the same. Instead maybe we should think of shadows a being relative to a given light source. Then you can still have the laser passing through a 3-d shadow that is crated by a light bulb. Actually...thinking about it more now, I see a shadow created by particular light source as a 3-d space in which if you palce an object anywhere in that space light from the source doesn't reach it. Even if the laser passing through this space hits the object the oblect is still in the shadow from the light source despite being lit up by the laser.

Maybe we are thinking of all shadows as being the same. Instead maybe we should think of shadows a being relative to a given light source. Then you can still have the laser passing through a 3-d shadow that is crated by a light bulb. Actually...thinking about it more now, I see a shadow created by particular light source as a 3-d space in which if you palce an object anywhere in that space light from the source doesn't reach it. Even if the laser passing through this space hits the object the oblect is still in the shadow from the light source despite being lit up by the laser.

Thanks. I think I get it now

Thanks for your hints...I always get confused over closure of addition and multiplication in ideals... How do these sound now? (1) Let F be a field, I an ideal of F such that there exists a non-zero r in I. (If there doesn't, then I={0} and we are done). For this r we have a multiplicative inverse, s in F such that sr=1=rs. Since I is closed under multiplication by elements of F 1=sr is in I. Since addition in an ideal is closed 1+1+...+1 is in I. Any element in F can be written as 1+1+...+1 for some number of 1s, thus any element of F is in I. By definition, any element of I is in F, thus the two are equal. (2) Let R be a commutative ring whose only ideals are {0} and R. Suppose R is not a field. Then there exists r in R such that there is no multiplicative inverse for r in R. There is an ideal generated by r; <r> = { pr : pis in R} (={rp : p is in R} since R commutative). So we have <r> = {0} or R (by definition of R). <r> = {0} iff r = 0, so suppose ris not 0. Then <r> must be R. But we have 1 in R, which implies that pr=1=rp for some p in R. That is, there exists a multiplicative inverse for r in R. This is a contadiction and hence R must be a field.

I need to show that the only ideals of a field are the trivial one {0} and f itself. I'm not sure where to begin, or if there are any little tricks to use. So far I've thought about picking an ideal and assume it has an element not equal to 0. (If it doesn't have such an element then it is {0} an we are OK.) With this non-zero element of the ideal I would then add it to a non-zero element of the field. This too is in the ideal. (Am I right in assuming that any ideal of ring an hence a field is a subring under addition?) Since a field has inverses I could pick such an element out of it above so r+f=1, where r is in the ideal and f in the field. Then we have 1 in the ideal. 1 generates the whole field. Thus the ideal is the whole field. Is this correct or can you spot any mistakes in it? I also have to show a commutative ring, R, whose only ideals are {0} and R is in fact a field....but I have had a go at that question yet.

Sorry about confusing you more Grantzam I solved the equations in a slightly different way to Bloodhound and I got the same solutions for a and b. I'm pretty sure they are the correct ones.