Let $G$ be an algebraic torus ($\cong(\mathbb C^*)^n$) acting linearly on a finite dimensional $\mathbb C$ vector space $V$. For a character $\chi$ of $G$, we define $V_{\chi}=\{v\in V|\forall g\in G, g\cdot v=\chi(g)v\}$.  I want to show that $V=\oplus_{\chi}V_{\chi}$.  I am following the book 'Linear algebraic groups' by Springer Theorem $3.2.3$, page $44$.  Let $X$ denotes the set of all characters of $G$ and $\phi:G\rightarrow GL(V)$ be the rational representation. This can be seen as a map to $\mathbb C^{n^{2}}$, therefore we have $\phi(g)_{i,j}\in\mathbb C[G]$ and we may write $\phi(g)_{i,j}=\sum_{\chi}a(i,j)_{\chi}\chi$, where $a(i,j)_{\chi}$ are constants. Thus we have \begin{align}\phi=\sum_{\chi}\chi A_{\chi}\end{align} for some $A_{\chi}\in GL(V)$(why $A_{\chi}$ is in $GL(V)$?). Note that only finitely many $A_{\chi}$ are non zero. Then by Dedikind's lemma we have $A_{\chi'}A_{\chi''}=\delta_{\chi',\chi''}A_{\chi'}$. We also have $\sum_{\chi}A_{\chi}=Id$. If we set $V_{\chi}=imA_{\chi}$, then $V=\oplus_{\chi}V_{\chi}$.  My question is why this $V_\chi:=imA_{\chi}=\{v\in V|\forall g\in G, g\cdot v=\chi(g)v\}$. In the book it is also given that $g\in G$ acts on $V_{\chi}$ as $\chi(g).Id$. I also don't understand this.  Any help is highly appreciated.  Thank you.
122488
Suppose a torus $T=(\mathbb{C}^\ast)^n$ acts on a finite dimensional vector space $W$, and define for $m \in M$ ($M$ is the character lattice of $T$) the eigenspace $W_m$ by  $$W_m = \{w \in W \mid t\cdot w = \chi^m(t)w \text{ for all }t \in T \}$$ i.e. for $w \in W_m$ is a simultaneous eigenvector for all $t \in T$, with eigenvalue $\chi^m(t)$ depending on $t \in T$. Then it is a famous fact $$W=\underset{m \in M} \bigoplus W_m$$ Can someone provide a somewhat self-contained proof of this result? I don't know much about the theory of algebraic groups.
54678
I'd like to see a proof why $\varphi \in \operatorname{Hom}{(S^1, S^1)}$ looks like $z^n$ for an integer $n$.  At first I thought I could argue that if I have a homomorphism that maps $e^{ix}$ to some $e^{iy}$ for $x,y \in \mathbb R$ then $z = rx$ for some real number $r$. But on second thought I'm not sure why I can't have $\varphi (e^{ix} ) = e^{g(x)i}$ for a $g$ other than $g(x) = \lambda x$.   I'm also interested in seeing different proofs. I'm sure there are several ways to prove this.  Thanks for your help.
80724
I'm attempting to come up with a first order language $L$ that is able to describe vector spaces over fields. I came up with a few sets of nonlogical symbols.  $Rs_L=\{Scal,Vec\}$ where $Scal,Vec$ are unary relation symbols which I interpret to be true or false depending on whether $x$ is a scalar or vector.  Also, $Fs_L=\{+,\cdot,\}$ which are the usual addition and multiplication for both vectors and scalars.  For constant symbols, $Cs_L=\{0,1,0_V\}$, where $0$ and $1$ are regular field scalars for the additive and multiplicative identities, respectively, and $0_V$ is the additive identity for vectors.   Using these, I was able to formulate axioms for scalar multiplication on a given structure, and I define addition between a scalar and a vector to simply always give the 0 vector and define multiplication between two vectors to always be the 0 vector. This gives a theory such that a structure models it if and only if it is a vector space in the regular algebraic sense. It is easy to see that all linear transformations in the algebraic sense are included in the class of homomorphisms (in the logical sense), but I am having difficulty proving they are the only ones in this class.   I suppose I want to show that for any $L$-homomorphism $T$, $T(av+w)=aTv+Tw$ for $a$ a scalar and $v,w$ vectors. Since $T$ is a homomorphism, I see that $T(av+w)=Tav+Tw=TaTv+Tw$, so it would be enough to show that all homomorphisms act as the identity on all scalars? Is there a way to show this, or an extra axiom to include?
77157
I am familiar with the interpretation of the irreducible representations (finite dimensional) of $SU(2)$ in terms of homogeneous polynomials of degree $n$.   If I take two of these irreducible representations, say $\rho_{i}$ and $\rho_{j}$ of dimensions $i+1$ and $j+1$ respectively, and take their product $\rho_{i}\rho_{j}$, the resulting representation will be the direct sum of other $\rho_{k}$'s. For a fixed $k$, how do I calculate the multiplicity of $\rho_{k}$ as a summand in the product representation in terms of $i$ and $j$?
121665
Let $V$ and $W$ be vector spaces (say over the reals). There is a linear injection $V^* \otimes W^* \to (V \otimes W)^*$ which sends $\sum_i f_i \otimes g_i \in V^* \otimes W^*$ to the unique functional in $(V \otimes W)^*$ sending $v \otimes w \mapsto \sum_i f_i(v) \cdot g_i(w)$ for all $(v,w) \in V \times W$. In the finite dimensional case it is easy to see this is an isomorphism by comparing dimensions on both sides. I'm aware that this is not an isomorphism in the infinite-dimensional case (see Relation with the dual space in the  on tensor products) but I must admit that I'm not sure why. Cardinal arithmetic does not seem help here and, even if it did, I would be much happier to see an explicit example of a functional in $(V \otimes W)^*$ outside the range of this map. I'm having trouble cooking one up myself.  Added: Let me elaborate on my comment about cardinal arithmetic. Suppose that $X$ and $Y$ are infinite dimensional vector spaces over a field $k$. I'm confident that $\dim (X \otimes Y) = \dim X \cdot \dim Y$ holds. It is clear that $|X^*| = |k|^{\dim X}$ since we may identify a functional on $X$ with a function from a basis for $X$ to $k$. Also I think I've convinced myself that if $\dim X \geq |k|$ then in fact $|X| = \dim X$ ie. the cardinality and dimension of a vector space agree when the dimension is larger than the cardinality of the ground field. Consequently, assuming say that $|k| \leq \dim X \leq \dim Y$ it seems we have $$ \dim(X \otimes Y)^* = |k|^{ \dim X \cdot \dim Y} = |k|^{ \dim Y} = \dim Y^* = \dim X^* \cdot \dim Y^* = \dim (X^* \otimes Y^*)$$  which implies that in fact $(X \otimes Y)^*$ and $X^* \otimes Y^*$ are isomorphic but, rather frustratingly, the obvious map does not do the job. Did I make a mistake here? If not, is it generally true (ie without making assumptions about $|k|$) that there exists some isomorphism $(X \otimes Y)^* \to X^* \otimes Y^*$?
37454
Good evening everybody,  I have a question about the red tagged argument in : Let $g$ be a semisimple Lie subalgebra of $gl(V )$.  For $x \in g$ we have J-C decomposion $ x= x_s +x_n$ but that's not clear to me why $ x_s ,x_n  \in n_{gl(V)}(g) $ where $  n_{gl(V)}(g) = \{y \in gl(V) | [y,x] \in g $ for every $ x \in g \} $.  My idea: We know that $ x_s = P(x) ,x_n = S(x) $ for appropriate polynomials without constant terms. Thats clear that $x \in n_{gl(V)} $ so because $ g$ is a subspace it would be enough to show that $[z,x^n] \in g $ for every $z \in g$. But how?
68576
Let $V$ be a representation with character $\chi$.  I would like to have a formula for the characters of the representations $\mathrm{Sym}^m[V]$ and $\wedge ^m[V]$ in terms of $\chi$.  Fulton and Harris presents such a formula for $m=2$, but not for general $m$.  Is there a "nice" such expression?  I found a paper by , but this doesn't seem to be what I'm looking for (it doesn't seem to give a general formula in terms of $\chi$).  In case the link does not work, here is the reference:  X. Zhou, P. Pulay. Characters for Symmetric and Antisymmetric Higher Powers of Representations: Application to  the  Number of Anharmonic Force Constants in Symmetrical Molecules. Journal of Computational Chemistry, Vol 10, Issue 7, (1989).
27264
I came across the claim  which states that there are no complex semisimple Lie algebras of dimension $4$, $5$, or $7$. As the problem suggests, we can take a Cartan subalgebra $H$ and root system $\Phi$ so that $L=H\oplus(\oplus_{\alpha\in\Phi}L_\alpha)$ where $$L_\alpha=\{x\in L:(\forall h\in H)\ [h,x]=\alpha(h)x\}$$  As shown in Humphreys' Introduction to Lie Algebras, every $L_\alpha$ has dimension $1$, so $\dim(L)=\dim(H)+\vert\Phi\vert$. I know $\Phi$ spans $H^*$, so $\dim(H)\leq\vert\Phi\vert$. Furthermore, I know that for any $\alpha\in\Phi$, the only multiples of $\alpha$ in $\Phi$ are $\alpha$ and $-\alpha$. Since $\mathbb{C}$ has characteristic $0$, this implies $\vert\Phi\vert$ is even.  Finally, another fact which may be useful is that given nonzero vectors $x_\alpha\in L_\alpha$ and $y_\alpha\in L_{-\alpha}$, if we take $h_\alpha=[x_\alpha,y_\alpha]\in H$, then $x_\alpha$, $y_\alpha$, and $h_\alpha$ span a three-dimensional subspace of $L$ isomorphic to $\mathfrak{sl}(2,\mathbb{C})$. However, I am not sure how to connect these pieces to give the desired conclusion.  As a follow-up question, would the claim still be true if I replaced the field with an arbitrary field of characteristic $0$, or is it also necessary for the field to be algebraically closed?
82698
The proof given by my textbook is highly non-satisfying. The author adopted some magic-like "reductio ad absurdum" and the proof (although is correct) didn't reveal the nature of this problem. I made my own effort into it and tried a different approach. Yet I can't finish it.  Let $\mathscr{F}$ be a commuting family in $M_n(\mathbb{C}^n)$, and $A\in\mathscr{F}$, then $A$ has $n$ eigenvalues. We pick one, say $\lambda$. Let $x$ be one of its eigenvector.  We can easily prove that, if $A$ has no other eigenvector with eigenvalue $\lambda$ that linearly independent with $x$, which means that $\{cx|c\in\mathbb{C}\}$ are the only vectors satisfying $Ax=\lambda x$, then $x$ is a common eigenvector. Because $\forall B\in\mathscr{F}$ and $\forall y\in \{cx|c\in\mathbb{C}\}$, $$A(Bx)=ABx=BAx=B(Ax)=B(\lambda x)=\lambda (Bx)$$, so that $Bx$ has to be in $\{cx|c\in\mathbb{C}\}$, that is, $Bx=c_0x$ for some $c_0\in\mathbb{C}$, which means $x$ is a eigenvector of $B$ too.  But what if there are vectors satisfying $Ax=\lambda x$ that's not in  $\{cx|c\in\mathbb{C}\}$? Well, then we should have a set of linearly independent eigenvectors $\{x_1,x_2,...,x_k\}$, that $\{c_1x_1+c_2x_2+...+c_kx_k|c_i\in\mathbb{C}\}$ are the only vectors satisfying $Ax=\lambda x$.  Now, I have a reasonable hypothesis that there exists some $x=c_1x_1+c_2x_2+...+c_kx_k$, that can be proven to be a common eigenvector of $\mathscr{F}$. I've tried some approaches to prove it but all failed.  Do you guys believe it's true? And if it is true then how do I prove it?
31985
Compute the fundamental group of the space obtained from two tori $S^1✕S^1$ by identifying a circle $S^1✕${$x_0$} in one torus with the corresponding circle $S^1✕${$x_0$} in the other.  Using van Kampen's theorem, I can fairly quickly show that if $T_1$ is one torus and $T_2$ is the other, then the group has to be isomorphic to $((π_1(T_1) ∗ π_1(T_2)))/N ≅ (\mathbb{Z} ✕ \mathbb{Z})∗ (\mathbb{Z} ✕ \mathbb{Z})/N$, where N is generated by elements of the form $i_{12}(w)i_{21}(w)^{-1}$, w is in $\pi_1(S^1✕${$x_0$}$) ≅ \mathbb{Z}$ and $i_{12}$, $i_{21}$ are the maps from $\pi_1(S^1✕${$x_0$}$)$ to $π_1(T_1)$, $π_1(T_2)$ respectively determined by inclusion. My problem is, as it seems to always be with van Kampen-related problems, figuring out what the elements of N look like. My attempt to do this in a pure-algebra way, as far as I can tell, failed me: I tried examining the case where w represents a loop that goes n (some integer) times around the circle, but then it seems like $i_{12}(w)$ &quot;$=$&quot; $(n, 0)$ in $T_1$ and $i_{21}(w)$ &quot;$=$&quot; $(n, 0)$ in $T_2$, which says to me that $i_{12}(w)i_{21}(w)^{-1} $&quot;$=$&quot;$ (n, 0) + (-n, 0) = 0$. The notation here isn't right, since $(n, 0)$ in $T_1$ is not the same as $(n, 0)$ in $T_2$ and their addition (composition?) shouldn't be written quite that way, hence my quotes around the equals signs, but certainly no loop in $S^1✕${$x_0$} is going to magically include anything but the 0 loop in either torus's second component, right? But this doesn't seem right to me, since if a, b are the generators of the circles in $T_1$ and d, e are the generators of the circles in $T_2$, and c is the constant path then the identification of the circle should make, say, $a = d$, so for example $(a, b)*(d, e)$ $ = (a^2,b)*(c, e)$ $ = (c, b)*(d^2,e)$. This doesn't match with what I just worked out N might be, but I also can't figure out what N should look like to make this equivalence make sense. My attempt to understand this geometrically has gone even worse. What I know for sure is that the resulting figure isn't the &quot;2-torus&quot;, 2 tori glued together at a disc. What I don't know for sure is whether this identification should actually create the hypertorus $S^1✕S^1✕S^1$ or something completely different.
54643
The twisted cubic is the image of the morphism $\phi : \mathbb{P}^1 \to \mathbb{P}^3 , (x:y) \mapsto (x^3:x^2 y:x y^2:y^3)$, it is given by $X = V(ad-bc,b^2-ac,c^2-bd)$. Now I would like to compute $I(X)$, which equals by the Nullstellensatz the radical of the ideal $I := (ad-bc,b^2-ac,c^2-bd) \subseteq k[a,b,c,d]$. I think that that $I$ is already a radical ideal, even a prime ideal. Namely, I suspect that  $$\phi^* : Q:=k[a,b,c,d]/I \to k[s,t] , a \mapsto s^3, b \mapsto s^2 t , c \mapsto s t^2 , d \mapsto t^3$$ is an injection. If this is true: How can we prove that? I've already tried to find a $k$-basis of the quotient, but this turned out to be a big mess. Even the representation of the quotient as a monoid algebra doesn't seem to help. Another idea is the following: A formal manipulation of generators and relations implies $Q_a \cong k[a,b]_a$. Thus it suffices to prove that $Q \to Q_a$ is injective, i.e. that $a$ is not a zero divisor.
51151
I computed the homology groups of the torus, can someone tell me if this is correct? The calculation, not the result that is. Thanks!  The cells of $T^2$ are $e^0, e^1_a, e^1_b, e^2$  The chain groups are  $$ C_0(T^2) = \{ k e^0 |&#160;k \in \mathbb{Z} \} = \mathbb{Z}$$  $$ C_1(T^2) = \{&#160;k_1 e^1_a + k_2 e^1_b |&#160;k_1 , k_2 \in \mathbb{Z} \} = \mathbb{Z} \oplus \mathbb{Z}$$  $$ C_2(T^2) = \mathbb{Z}$$  $$ C_k(T^2) = \{ 0 \} , k &gt; 2$$  Now the homology groups:  $$ H_0(T^2) = \ker \partial_0 / im \partial_1 = \mathbb{Z} / 0 = \mathbb{Z}$$  where $im \partial_1 = 0$ because there is no chain in $C_1(T^2)$ whose boundary is a zero-chain in $C_0(T^2)$. (Is this reasoning correct?)  $$ H_1(T^2) = \ker \partial_1 / im \partial_2 = \mathbb{Z} \oplus \mathbb{Z}$$  where $\ker \partial_1 =  \mathbb{Z} \oplus \mathbb{Z} $, i.e. again everything maps to zero because there is no element in $C_1$ whose boundary maps to an element in $C_0$.  $im \partial_2 = 0$ again because there is no element in $C_2$ whose boundary is an element of $C_1$.  I don't want to use Hurewicz to do $H_1$.  $$ H_2(T^2) = \mathbb{Z}$$ using similar arguments as above.  Thanks for your help.  Edit  I posted the computations as an answer below. I got two up votes but I don't know by who so I'm not yet sure I can trust them...
121038
Let $V$ be a vector space over a field $F$ (not necessarily algebraically closed), and $T:V\rightarrow W$ be a $F$-linear map. If $c(T)$ and $m(T)$ denote the characteristic and minimal polynomial respectively, then I want to show that $c(T)$ and $m(T)$ have the same set of irreducible factors.  I know that the claim is true if $F$ is algebraically closed, because then the two polynomials have the same roots. The typical reasoning one gives when $F$ is not algebraically closed is: Take an algebraic closure of $F$, say $K$, and look at $m(T)$ and $c(T)$ in $K$ to conclude the proof. But as far as I know, the proof that any root of $c(T)$ is a root of $m(T)$ requires the fact that $V$ is a $K$-vector space. But we only know that $V$ is an $F$-vector space. I can't understand how to proceed after this.
92801
I want to ask about the homeomorphism between $\mathbb{Q}$, $\mathbb{Q}_{&gt;0}$: the rationals greater than $0$, and $\mathbb{Q}_{\geqslant 0}$: the rationals $\geqslant 0$?  For $\mathbb{Q}$ and $\mathbb{Q}_{&gt;0}$, I can use the back and forth map to have an order-isomorphism, and as a result, a homeomorphism. Is there a direct formula for the map?  It's very surprising that $\mathbb{Q}_{\geqslant 0}$ is homeomorphic to $\mathbb{Q}_{&gt;0}$ and $\mathbb{Q}$,  I don't know how to show they are homeomorphic.    Could anyone please help me with the homeomorphisms between them?
114947
Let $\mathbb{T}^2$ be the torus and have $\mathbb{Z}_2$ action on the torus by permuting the coordinates. I am trying to prove the orbit space is congruent to the mobius strip $\mathbb{T}^2/\mathbb{Z}^2 \cong M$.  I am writing $M\cong I^2/\alpha$, where $I=[0,1]$, $\alpha(x)=1-x$ and we identify the points $(x,0)\sim (\alpha(x),1)$. I know it is useful for this problem to identify the torus as the quotient $I^2/\sim_0$ with left glued to right and top glued to bottom (referring to sides) and then the action on the torus corresponds to switching coordinates on the square, which is reflecting over the diagonal $(x,x)$. What I don't understand is how to combine these relations of gluing and permuting to show this space is homeomorphic to the quotient above with $\alpha$.   I think that with both relations we will identify the edges together since $(x,0)\sim_0(x,1)\sim(1,x) \sim_0(0,x)$. Then points not on the edges will either be identified with themselves if they are on the diagonal or just their permutation. But I can't figure out how to map this to the Mobius strip...  How do you proceed formally with proving these spaces are homeomorphic? Obviously I'm not looking for all the details but I'm stuck both trying to visualize it (i get to folding the square over the diagonal $(x,x)$ but then I cant picture how to get the mobius strip from there on). Any help is appreciative.   Also a different generalization than the natural one: instead of permuting the coordinates in $\mathbb{T}^k$ and taking the action by $S_n$ consider the action by reversing the coordinates, which can still be done by using $\mathbb{Z}_2$. What is this orbit space homeomorphic to? Would it be $M^k$ when $k$ is even and something else for odd $k$?
93722
For an algebraic geometry assignment I recently had to prove that the isomorphism classes of line bundles over $\mathbb{P}_k^n$ for any field $k$ are the powers of the tautological line bundle, or in other words the Picard group is given by Pic($\mathbb{P}_k^n) \cong \mathbb{Z}$.  I was trying to consider simple examples to reconcile this with my intuition about line bundles from algebraic topology.  Taking $k = \mathbb{R}$ and $n = 1$, we have Pic($\mathbb{RP}^1) \cong \mathbb{Z}$.  Naively, I might have thought that since $\mathbb{RP}^1 \cong S^1$, there would be only the two isomorphism classes of (topological) line bundles over the circle.  I was wondering: what are the relations between line bundles in algebraic geometry and topology in this case, and why the discrepancy?    I figure maybe the real numbers are not a very well behaved field in algebraic geometry. If $k = \mathbb{C}$ and we again take $n = 1$, do the classifications of bundles match up on $\mathbb{CP}^1 \cong S^2$ in the geometric and topological contexts?
91930
I'm currently reading Goldblatt's Topoi - The categorial analysis of logic and at some point he lets one verification to the reader and I must admit that I'm stuck.  He's just introduced $\mathbf{M}$-sets where $\mathbf{M}$ is a monoid, and he's showing that the category of $\mathbf{M}$-sets and multiplication-preserving maps is a topos.  He also introduced the concept of a left ideal and called $L_M$ the set of left ideals of $M$, and his claim is that $L_M$ together with the multiplication $m\cdot B = \{n\mid nm \in B\}$; and with the true arrow $true : \{0\} \to L_M$, $true(0) = M$ is a subobject classifier in this category.   Then he considers a monic arrow $k: X\to Y$ and argues that the character $\chi : Y\to L_M$ associated with this subobject is defined by $\chi(y) = \{n \mid n\cdot y \in X\}$.   This is where I'm stuck : I've managed to show that this does make the associated square a pullback square, and that any other arrow $g:Y\to L_M$ that makes it a pullback square has $g(x) = M$ for $x\in X$. I've also shown that it has $\chi(y) \subset g(y)$ for all $y\in Y$, but I can't prove that $\chi$ is the only one that works...  Any indications?