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7

1. In the general case, Weibel's Exercise 7.3.2 is wrong: The Lie subalgebra $\mathfrak{f}\left( M\right) $ of $\operatorname{Lie}\left( T\left( M\right) \right) $ is not always a free Lie algebra on $M$ (at least not if the inclusion of $M$ into it is considered as the canonical map). To see why not, we shall use the counterexample constructed in §5 of ...


6

If $G$ is compact and connected, one can prove (by constructing a bi-invariant metric on $G$ and relating the metric and Lie group exponential maps) that the exponential map $\exp : {\frak g} \to G$ is surjective. This justifies the claim for, e.g., $SU(2)$. Using the Baker-Campbell-Hausdorff formula, one can show that $\exp$ is also surjective for Lie ...


5

Let $f:G\to H$ be an injective Lie group morphism with $H$ compact. Since $H$ has a faithful finite dimensional representation, it suffices to compose with $f$ to obtain one for $G$. This shows that any Lie group $G$ which does not have faithful finite dimensional representations provides an example of what you want.


4

For characteristic $2$, the Lie algebra $\mathfrak{sl}_2(F)$ is the Heisenberg Lie algebra, with basis $(x,y,z)$ and Lie bracket $[x,y]=z$. Hence its commutator subalgebra is $1$-dimensional, i.e., not equal to itself. For characteristic different from $2$, a computation with elementary matrices $E_{ij}$ shows the result, i.e., the computation of the ...


3

The fact that $i\circ\phi$ is isomorphic to $i$ doesn't imply that $\phi$ is an inner automorphism.


3

$\text{ad}([X, Y]) = [\text{ad}(X), \text{ad}(Y)]$, so from here it suffices to prove that $\text{tr}([M, N]) = 0$ for any square matrices $M$ and $N$. This is equivalent to saying that $\text{tr}(MN) = \text{tr}(NM)$, which is a nice exercise. You can prove it by writing everything out explicitly but there are also nicer proofs. If $\mathfrak{g}$ weren't ...


3

No. Consider the adjoint rep of sl3: the outer weightspaces are 1-d and the zero weightspace is 2d so there is no way the map induced by the action of $e_\alpha$ from $M_{-\alpha}$ to the zero weight space can be onto. – mt_ 5 hours ago


3

Let's start with the first example and use your $i,j,k$ for the $SO(2,1,R)$ matrices. The analogous $SL(2,R)$ matrices with the same commutators are multiples of the Pauli matrices $$ j = \frac 12 \pmatrix{ 0&1\\1&0 },\,\, i = \frac 12 \pmatrix{ 0&-1\\1&0 },\,\, k = \frac 12 \pmatrix{ 1&0\\0&-1 }, $$ You may easily verify that these ...


3

The Killing form of $\mathfrak{gl}_n(K)$ restricts to the Killing form on its ideals, but not on its subalgebras. Since $\mathfrak{sl}_n(K)$ is an ideal in $\mathfrak{gl}_n(K)$, the Killing form of $\mathfrak{sl}_n(K)$ really is $B(x,y)=2n\;tr(xy)$. The Killing form for $\mathfrak{sp}_{2n}(K)$ is given by $B(x,y)=(4n+2)\, tr(xy)$, and for ...


3

As you say, it depends on context. Vaguely, it means something like "every element except for the elements in a 'small' subset," where the meaning of 'small' depends on context. In this context it might mean either every element except for the elements in a finite (or maybe countably infinite) union of affine subspaces, or every element except for the ...


2

This is almost tautological. "Bi-invariant" means the metric is invariant under both left and right translation. The resulting distance function satisfies, in particular, $$d(gx,gy)=d(x,y)$$ Hence left translation is an isometry (as is right translation).


2

Let the entry in row $j$, column $k$ be $$ u_{jk} = e^{2\pi i j k/n}$$ This is related to the properties of the finite Fourier transform. If you want your $u_{jk}$ to be real, you're looking for Hadamard matrices. These can only exist if $n$ is $1$, $2$ or a multiple of $4$. The existence of such matrices for all multiples of $4$ is the Hadamard ...


2

The Killing form on a Lie algebra doesn't restrict to the Killing form on its subalgebras. When you take the trace needed to compute the Killing form it's over a smaller and different vector space.


2

The following uses the theory of Jordan decomposition in finite dimensional semisimple complex Lie algebras (namely in $\mathfrak{sl}_2(\mathbb{C})$) and some basic knowledge about the finite dimensional representations of $\mathfrak{sl}_2(\mathbb{C})$. There is probably a much shorter and easier way to do this, but it’s the best I’m coming up with right ...


2

There are a couple of things "generate" could mean here. The weakest reasonable one, which suffices in this case, is "generates under taking tensor products, finite direct sums, and direct summands." In general, depending on the situation, you might want to take other kinds of colimits. For references try looking up anything that deals with Tannaka ...


2

Presumably you understand from the rest of the question that the $H$-eigenspace decomposition of $W=V_{\lambda}$ is $W=\bigoplus_{n=0}^\lambda W(\lambda-2n)$ with each eigenspace, $W(\lambda-2n)$, being 1-dimensional. By part (i) of the question together with the decomposition above, you know that the $\lambda$-eigenspace of $V/W\cong nV_\lambda$ is ...


2

The standard definition of the Lie bracket is: For a differentiable function $f$ on $M$ and two vector fields $X, Y$ on $M$, $$ [X,Y](f)= XY(f) - YX(f). $$ You need to check that this defines a derivation on the space of functions on $M$. Doing so is a very nice exercise in the definitions (the more interesting part is verifying the Leibnitz rule).


2

Gabriel's theorem is the one involving finite root systems in the representation theory of quivers. It basically says that a connected quiver admits finitely many indecomposable representations if and only if it is an orientation of a Dynkin diagram of type ADE, and gives a bijection between the set of indecomposable representations and the set of positive ...


2

Your guess is correct, with the exception that also the center is always a Lie ideal. Two days ago we have determined the center of $\mathfrak{gl}_n(K)$ here, for a field $K$ of characteristic zero. Indeed, the center is $1$-dimensional, spanned by $I_n$. Since $\mathfrak{gl}_n(K)\cong \mathfrak{sl}_n(K)\oplus K$, and $\mathfrak{sl}_n(K)$ is semisimple, the ...


2

If $\varphi : G \to G$ is an automorphism of $G$, then given any representation $\rho : G \to \text{Aut}(V)$, we get another representation $\rho \circ \varphi : G \to \text{Aut}(V)$, which will generally be different if $\varphi$ is an outer automorphism. In particular the two $3$-dimensional representations of $SL_3$ are related by the nontrivial outer ...


2

If you suppose that $G$ is also compact then you have a left invariant metric on $G$. Let $p:G\rightarrow G/H$ the projection and $x$ be a point of $G$, you can identify the tangent space of $p(x)$ to the orthogonal $N_x$ of the subspace of $T_xG$ tangent to the orbit $H(x)$. The restriction of the invariant metric on $N_x$ defines a metric on $G/H$.


2

This happens already for $\mathfrak{sl}_2$: The only non-semisimple block (up to equivalence) of BGG category $\mathfrak{sl}_2$ has two simple modules, $L$ and $S$ with $\operatorname{dim} \operatorname{Ext}^1(L,S)=\operatorname{dim}\operatorname{Ext}^1(S,L)=\operatorname{dim}\operatorname{Ext}^2(S,S)=1$. In this case, the category has global dimension $2$, ...


1

Let $G$ be a connected Lie group and $H$ a closed subgroup. The set of $G$-invariant Riemannian metrics on $G/H$ is in canonical bijection with the set of $H$-invariant scalar products on $\mathfrak{g}/\mathfrak{h}$ under the adjoint representation of $G$. If $H$ is compact, the latter is non-empty. If $H$ is trivial, this set of metrics is thus quite big ...


1

I am not 100% sure it would match your expectations but have a look at Peter Olver's Applications of Lie Groups to Differential Equations. Specifically for the mathematics surrounding the Noether theorem you also may wish to look at the book by Yvette Kosmann-Schwarzbach, The Noether theorems. Invariance and conservation laws in the twentieth century, ...


1

1.) We have $ad(L)=ad([L,L])=[ad(L),ad(L)]$, hence every $ad(x)$ can be written as a linear combination of elements $[ad(y),ad(z)]$, which all have trace zero. Hence $tr(ad(x))=0$ for all $x\in L$. 2.) $L$ cannot be solvable because of $[L,L]=L$. Now suppose that $I$ is a nonzero ideal, different from $L$. Hence $dim (I)=1$ or $dim(I)=2$. In the first case, ...


1

Consider the adjoint representation. Since the algebra is perfect, we have $tr\ ad(x)=0$ for all $x \in L$. Since $ad(x)$ has determinant zero, because $[x,x]=0$, we see that the characteristic polynomial of $ad(x)$ is of the form $f(t) = t^3 + c(x) t$ for some scalar $c(x)$ depending on $x$. If we had $c(x) = 0$ for all $x$, then $ad(x)$ would be ...


1

Let $(e_1,e_2,e_3)$ be a basis of $L$. Since $[L,L]=L$, $$ (f_1,f_2,f_3):=([e_2,e_3],[e_3,e_1],[e_1,e_2]) $$ is again a basis of $L$. Hence we may write the $f_i$ as linear combinations of the $e_i$, i.e., $f_i=\sum a_{ij}e_j$, with invertible coefficient matrix $A=(a_{ij})$. One checks that $A$ must be symmetric, and that every invertible symmetric matrix ...


1

In $\mathbb{R}^3$, we can indeed more or less identify $u\times v$ with $u\wedge v$ for two vectors $u,v \in \mathbb{R}^3$ (more accurately, $u\wedge v$ is the linear form given by scalar product with $u\times v$; or another way of putting thigs it is that $u\wedge v$ gives an "axial vector" or "pseudovector", not a vector), but this no longer works for ...


1

Here is the proof from the paper by Tôgô (1967) mentioned by Dietrich. It works in arbitrary characteristic, for arbitrary nonzero nilpotent Lie algebras (possibly of infinite dimension). Let $K$ be the ground field. Let $\mathfrak{g}$ be a nonzero nilpotent Lie algebra. Write it as semidirect product $K\ltimes\mathfrak{h}$. If $\dim(\mathfrak{g})$ is ...


1

I am assuming you are dealing with real or complex Lie Algebras, not on an arbitrary field. If $H$ is compact, you can define an $\mathrm{Ad}_H$-invariant inner product on $\mathfrak g$. Since $\mathfrak h$ is a $\mathrm{Ad}_H$-invariant subspace, its orthogonal complement (with respect to this inner product) will also be $\mathrm{Ad}_H$-invariant. Taking ...



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