Exercise 1.1
(i) Show that $[v, 0] = 0 = [0, v]$ for all $v \in L$
Solution
Since the Lie bracket is bilinear, we have

$$[v, 0] = [v, 0 + 0] = [v, 0] + [v, 0]$$

and

$$[0, v] = [0 + 0, v] = [0, v] + [0, v]$$

Subtracting $[v, 0]$ from the first equation and $[0, v]$ from the second gives the desired result. $\square$

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(ii) Suppose that $x, y \in L$ satisfy $[x, y] \neq 0$. Show that $x$ and $y$ are linearly independent over $F$.
Solution
The contrapositive of the statement is to assume that $x$ and $y$ are linearly dependent over F and prove that $[x, y] = 0$. We shall prove the contrapositive. So take $y=kx$ for some $k \in F$. Then

$$[x, y] = [x, kx] = k [x, x] = 0$$

Since $[x, x] = 0$ by property (L1). $\square$

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Exercise 1.2
Let $F = \mathbf{R}$. The vector product $(x, y) \mapsto x \wedge y$ defines the structure of a Lie algebra on $\mathbf{R}^3$. We denote this Lie algebra by $\mathbf{R}^3_\wedge$. Explicitly, if $x=(x_1, x_2, x_3)$ and $y=(y_1, y_2, y_3)$ then

$$x \wedge y = (x_2y_3 - x_3y_2, x_3y_1 - x_1y_3, x_1y_2 - x_2y_1)$$

Convince yourself that $\wedge$ is bilinear. Then check that the Jacobi identity holds. Hint: If $x \cdot y$ denotes the dot product of the vectors $x, y \in \mathbf{R}^3$, then

$$x \wedge (y \wedge z) = (x \cdot z)y - (x \cdot y)z \quad \text{for all } x,y, z \in \mathbf{R}^3$$

Solution
Let $a, b \in \mathbb{R}$ and $x, y, z \in \mathbf{R}^3$. To show that $\wedge$ is bilinear we must show that

$$(a x + b y) \wedge z = a (x \wedge z) + b (y \wedge z), \\ z \wedge (a x + b y) = a ( z \wedge x) + b(z \wedge y),$$

So

$$\begin{aligned} (a x + b y) \wedge z &= (ax_1 + by_1, ax_2 + by_2, ax_3 + by_3) \wedge (z_1, z_2, z_3) \\ &= \big((ax_2 + by_2)z_3 - (ax_3 + by_3)z_2, \\ &\quad (ax_3 + by_3)z_1 - (ax_1 + by_1)z_3, \\ &\quad (ax_1 + by_1)z_2 - (ax_2 + by_2)z_1\big) \\ &= \big(a(x_2z_3 - x_3z_2) + b(y_2z_3 - y_3z_2), \\ &\quad a(x_3z_1 - x_1z_3) + b(y_3z_1 - y_1z_3), \\ &\quad a(x_1z_2 - x_2z_1) + b(y_1z_2 - y_2z_1)\big) \\ &= a(x_2z_3 - x_3z_2, x_3z_1 - x_1z_3, x_1z_2 - x_2z_1) \\ &\quad + b(y_2z_3 - y_3z_2, y_3z_1 - y_1z_3, y_1z_2 - y_2z_1) \\ &= a (x \wedge z) + b (y \wedge z) \end{aligned}$$

To show that $z \wedge (a x + b y) = a ( z \wedge x) + b(z \wedge y)$ we could repeat the above but reversed, but there is a shortcut. Calculate that

$$\begin{aligned} y \wedge x &= (y_2x_3 - y_3x_2, y_3x_1 - y_1x_3, y_1x_2 - y_2x_1) \\ &= - (y_3x_2 - y_2x_3, y_1x_3 - y_3x_1, y_2x_1 - y_1x_2) \\ &= - (x_2y_3 - x_3y_2, x_3y_1 - x_1y_3, x_1y_2 - x_2y_1) \\ &= - x \wedge y \end{aligned}$$

So

$$z \wedge (a x + b y) = - ((a x + b y) \wedge z) = -(a (x \wedge z) + b (y \wedge z)) = a(z \wedge x) + b(z \wedge y)$$

Finally we must show that the Jacobi identity holds for $\wedge$. This is easy using the given identity

$$x \wedge (y \wedge z) + y \wedge (z \wedge x) + z \wedge (x \wedge y) \\= ((x \cdot z)y - (x \cdot y)z) + ((y \cdot x)z - (y \cdot z)x) + ((z \cdot y)x - (z \cdot x)y) = 0. \quad \square$$

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Exercise 1.3
Suppose that $V$ is a finite-dimensional vector space over $F$. Write $\mathrm{gl}(V)$ for the set of all linear maps from $V$ to $V$. This is again a vector space over $F$, and it becomes a Lie algebra, known as the general linear algebra, if we define the Lie bracket $[-, -]$ by

$$[x, y] \coloneqq x \circ y- y \circ x \quad \text{for } x, y \in \mathrm{gl}(V),$$

where $\circ$ denotes the composition of maps. Check that the Jacobi identity holds.
Solution
Calculate that

$$\begin{aligned} &\phantom{=}[x, [y, z]] + [y, [z, x]] + [z,[x,y]] \\ &= x \circ [y, z] - [y, z ]\circ x \\ &\quad + y \circ [z, x] - [z, x ] \circ y \\ &\quad + z \circ [x, y] - [x, y] \circ z \\ &= x \circ (y \circ z - z \circ y) - (y \circ z - z \circ y) \circ x \\ &\quad + y \circ (z \circ x - x \circ z) - (z \circ x - x \circ z) \circ y \\ &\quad + z \circ (x \circ y - y \circ x) - (x \circ y - y \circ x) \circ z. \\ &= x \circ y \circ z - x \circ z \circ y - y \circ z \circ x + z \circ y \circ x \\ &\quad + y \circ z \circ x - y \circ x \circ z - z \circ x \circ y + x \circ z \circ y \\ &\quad + z \circ x \circ y - z \circ y \circ x - x \circ y \circ z + y \circ x \circ z \\ &= 0 \end{aligned}$$

It seems that all the terms cancelled out nicely! I see why this exercise is famous as one that every mathematician should do at least once in their life 😃 $\square$

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Unnamed Exercise
Write $\mathrm{gl}(n, F)$ for the vector space of all $n \times n$ matrices over $F$ with the Lie bracket defined by

$$[x, y] \coloneqq xy - yx$$

Where $xy$ is the usual product of the matrices $x$ and $y$.
As a vector space, $\mathrm{gl}(n, F)$ has a basis consisting of the matrix units $e_{ij}$ for $1 \leq i, j \leq n$. Here $e_{ij}$ is the $n \times n$ matrix which has 1 in the $ij$-th position and all other entries are 0. We leave it as an exercise to check that

$$[e_{ij}, e_{kl}] = \delta_{jk}e_{il} - \delta_{il}e_{kj}$$

where $\delta$ is the Kronecker delta, defined by $\delta_{ij} = 1$ if $i = j$ and $\delta_{ij} = 0$ otherwise.
Solution

$$[e_{ij}, e_{kl}] = e_{ij}e_{kl} - e_{kl}e_{ij}$$

Using the formula for matrix multiplication at some index $1 \leq a, b \leq n$ we get

$$(e_{ij}e_{kl})_{ab} = \sum_{x = 0}^{n} (e_{ij})_{ax}\cdot(e_{kl})_{xb}$$

Whenever $x \neq j$ by definition we have $(e_{ij})_{ax} = 0$ so this reduces to

$$(e_{ij}e_{kl})_{ab} = (e_{ij})_{aj}\cdot(e_{kl})_{jb} = \delta_{ai} \cdot \delta_{kj} \cdot \delta_{lb}$$

Hence we have

$$e_{ij}e_{kl} = \delta_{jk} \cdot e_{il}$$

As by definition $(e_{il})_{ab} = \delta_{ai} \cdot \delta_{lb}$. Rearranging symbols gives $e_{kl}e_{ij} = \delta_{li}e_{kj}$, combining the results finishes the proof. $\square$

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Exercise 1.4
Check the following assertions:
Let $\mathrm{b}(n,F)$ be the upper triangular matrices in $\mathrm{gl}(n,F)$. (A matrix $x$ is said to be upper triangular if $x_{ij}=0$ whenever $i > j$.) This is a Lie algebra with the same Lie bracket as $\mathrm{gl}(n,F)$.

Similarly, let $\mathrm{n}(n,F)$ be the strictly upper triangular matrices in $\mathrm{gl}(n,F)$. (a matrix $x$ is said to be strictly upper triangular if $x_{ij}=0$ whenever $i \geq j$). Again this is a Lie algebra with the same Lie bracket as $\mathrm{gl}(n,F)$.
Solution
Let $x, y$ be upper-triangular matrices. Recall from linear algebra that upper triangular matrices and strictly upper triangular matrices are subspaces of the vector space formed by all square matrices. Therefore, $(xy - yx)$ forms a Lie Algebra over $\mathrm{b}(n,F)$ and $\mathrm{n}(n,F)$, because the required properties of being a bilinear map $[:]:L \to L$ satisfying the properties

TODO You need to show that when you multiply two (strictly) upper triangular matrices you get a (strictly) upper triangular matrix!

$$\tag{L1} \phantom{(L1)} \quad [x,x]=0 \quad\text{for all } x \in L,$$

$$\tag{L2} \phantom{(L2)} \quad [x,[y,z]] + [y,[z,x]] + [z,[x,y]]=0 \quad \text{for all } x,y,z \in L.$$

are inherited from $\mathrm{gl}(n,F)$. $\quad \square$

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Exercise 1.5
Find $Z(L)$ when $L = \mathrm{sl}(2,F)$. You should find that the answer depends on the characteristic of $F$.
Solution
First let's just plug in

$$Z(\mathrm{sl}(2,F)) = \{ x \in \mathrm{sl}(2,F): xy - yx = 0 \text{ for all } y \in \mathrm{sl}(2,F) \}$$

Now since $x,y \in \mathrm{sl}(2,F)$ we know that their trace is zero

$$x = \begin{pmatrix} x_{11} & x_{12} \\ x_{21} & -x_{11} \end{pmatrix}, y = \begin{pmatrix} y_{11} & y_{12} \\ y_{21} & -y_{11} \end{pmatrix} \\[4pt] xy = \begin{pmatrix} x_{11} & x_{12} \\ x_{21} & -x_{11} \end{pmatrix} \cdot \begin{pmatrix} y_{11} & y_{12} \\ y_{21} & -y_{11} \end{pmatrix} = \begin{pmatrix} x_{11}y_{11} + x_{12}y_{21} & x_{11}y_{12} - x_{12}y_{11} \\ x_{21}y_{11} - x_{11}y_{21} & x_{21}y_{12} + x_{11}y_{11} \end{pmatrix} \\[4pt] yx = \begin{pmatrix} y_{11} & y_{12} \\ y_{21} & -y_{11} \end{pmatrix} \cdot \begin{pmatrix} x_{11} & x_{12} \\ x_{21} & -x_{11} \end{pmatrix} = \begin{pmatrix} y_{11}x_{11} + y_{12}x_{21} & y_{11}x_{12} - y_{12}x_{11} \\ y_{21}x_{11} - y_{11}x_{21} & y_{21}x_{12} + y_{11}x_{11} \end{pmatrix} \\[4pt] xy - yx = \begin{pmatrix} x_{12}y_{21} - y_{12}x_{21} & 2x_{11}y_{12} - 2x_{12}y_{11} \\ 2x_{21}y_{11} - 2x_{11}y_{21} & x_{21}y_{12} - y_{21}x_{12} \end{pmatrix}$$

So when is $xy - yx = 0$? Can we come up with some generators for this set? Keep in mind we are only picking for $x$, $y$ can be anything.

$$x_{12}y_{21} = y_{12}x_{21} \\ 2x_{11}y_{12} = 2x_{12}y_{11} \\ 2x_{21}y_{21} = 2x_{11}y_{21} \\ x_{21}y_{12} = y_{21}x_{12}$$

Let's look at some examples. Let F be the field of two elements. Then the only requirements for the center become

$$x_{12}y_{21} = y_{12}x_{21} \\ x_{21}y_{12} = y_{21}x_{12}$$

So consider the matrix

$$x = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$$

This matrix solves the equations above, to show it is in the center explicitly we can show

$$\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \cdot \begin{pmatrix} y_{11} & y_{12} \\ y_{21} & -y_{11} \end{pmatrix} = \begin{pmatrix} y_{11} & y_{12} \\ -y_{21} & y_{11} \end{pmatrix} \\[4pt] \begin{pmatrix} y_{11} & y_{12} \\ y_{21} & -y_{11} \end{pmatrix} \cdot \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} = \begin{pmatrix} y_{11} & -y_{12} \\ -y_{21} & y_{11} \end{pmatrix} \\[4pt] xy - yx = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \cdot \begin{pmatrix} y_{11} & y_{12} \\ y_{21} & -y_{11} \end{pmatrix} - \begin{pmatrix} y_{11} & y_{12} \\ y_{21} & -y_{11} \end{pmatrix} \cdot \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}\\[4pt] = \begin{pmatrix} y_{11} & y_{12} \\ -y_{21} & y_{11} \end{pmatrix} - \begin{pmatrix} y_{11} & -y_{12} \\ -y_{21} & y_{11} \end{pmatrix} = \begin{pmatrix} 0 & y_{12} + y_{12} \\ 0 & 0 \end{pmatrix}$$

Since for all $\lambda \in F_2$ we have $\lambda + \lambda = 0$, in $F_2$ we clearly have $xy - yx = 0$, so $x \in Z(\mathrm{sl}(2, F_2))$. Of course, we always have $0 \in Z(\mathrm{sl}(2,F))$ for all fields $F$, and when the characteristic of $F$ is not 2, this is the only element. $\quad \square$

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Exercise 1.6
Show that if $\varphi : L_1 \to L_2$ is a homomorphism, then the kernel of $\varphi$, $\ker \varphi$, is an ideal of $L_1$, and the image of $\varphi$, $\text{im} \varphi$, is a Lie subalgebra of $L_2$.
Solution
First let's show that $\ker \varphi$ is an ideal of $L_1$. The definition of a kernel gives us

$$\ker \varphi = \{x \in L_1: \varphi(x)=0 \}.$$

We wish to show that $\ker \varphi$ is an ideal of $L_1$.
First we must show that it is a subspace of $L_1$. This amounts to showing that $\ker \varphi$ contains the zero vector, and is closed under both vector addition and scalar multiplication. $0 \in \ker \varphi \iff \varphi(0) = 0$, the latter following from $\varphi$ being a homomorphism, so $\ker \phi$ contains the zero vector. Let $x, y \in \ker \varphi$ and $c \in F$ (where $F$ is the common field of scalars shared by $L_1$ and $L_2$). Then $\varphi(x + y) = \varphi(x) + \varphi(y) = 0$ and $\varphi(c \cdot x) = c \cdot \varphi(x) = 0$. We conclude that $\ker \varphi$ is a subspace of $L_1$
Let $x \in L_1$ and $y \in \ker \varphi$. To show that $\ker \varphi$ is an ideal of $L_1$, we need to show that $[x, y] \in \ker \varphi$. Calculate that

$$\varphi ([x,y]) = [\varphi(x), \varphi(y)] = [\varphi(x), 0] = 0 \quad \square$$

Next we show that $\text{im} \varphi$ is a Lie subalgebra of $L_2$. Recall that

$$\text{im} \varphi = \{ y \in L_2: y = \varphi(x), \text{ for some } x \in L_1 \}.$$

To prove that $\text{im} \varphi$ is a Lie subalgebra of $L_2$, we first need to show that $\text{im} \varphi$ is a vector subspace of $L_2$. Since $\varphi(0)=0$, $0 \in \text{im} \varphi$. If $y_1 = \varphi(x_1), y_2 = \varphi(x_2), c \in F, x_1, x_2 \in L_1$, then $y_1 + y_2 = \varphi(x_1) + \varphi(x_2) = \varphi(x_1 + x_2)$ and $cy_1 = c \varphi(x_1) = \varphi(cx_1)$, so $\text{im} \varphi$ is closed under vector addition and scalar multiplication. We conclude that $\text{im} \varphi$ is a vector subspace of $L_2$.
To show that $\text{im} \varphi$ is a Lie subalgebra of $L_2$, it remains to show that

$$[y_1, y_2] \in \text{im} \varphi \qquad \text{for all } y_1,y_2 \in \text{im} \varphi$$

So let $y_1 = \varphi(x_1), y_2 = \varphi(x_2)$. Calculate that

$$[y_1, y_2] = [\varphi(x_1), \varphi(x_2)] = \varphi([x_1, x_2]) \qquad \square$$

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Exercise 1.7
Let $L$ be a Lie algebra. Show that the Lie bracket is associative, that is, $[x, [y, z]] = [[x,y],z]$ for all $x, y, z \in L$, if and only if for all $a, b \in L$ the commutator $[a, b]$ lies in $Z(L)$.
Solution
First note that

$$\quad [a,b] \in Z(L) \text{ for all } a,b \in L \iff [[a,b],c] = 0 \text{ for all } a,b,c \in L$$

Then calculate that

$$[[a,b],c] = 0 \text{ for all } a,b,c \in L \implies [x,[y,z]] = [[x,y],z] = 0$$

and that

$$[x, [y, z]] = [[x,y],z] \text{ for all } x,y,z \in L \\ \implies [[x,y],z] = [x, [y,z]] = -[[y,z],x] = -[y, [z,x]] = [[z,x],y] = [z,[x,y]] = -[[x,y],z] \\ \implies [[x,y],z] = -[[x,y],z] \\ \implies 1 + 1 = 0 \text{ or } [[x,y],z] = 0 \quad \square$$

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Exercise 1.8
Let $D$ and $E$ be derivations of an algebra $A$.
(i) Show that $[D,E] = D \circ E - E \circ D$ is also a derivation.
Solution
Let $a,b \in A$, by the definition of a derivation we have

$$D(ab) = aD(b) + D(a)b \\ E(ab) = aE(b) + E(a)b$$

so

$$[D, E](ab) = D(E(ab)) - E(D(ab)) = D(aE(b)+E(a)b) - E(aD(b) + D(a)b) \\ = D(aE(b)) + D(E(a)b) - E(aD(b)) - E(D(a)b)$$

We need to show that

$$[D,E](ab) = a[D,E](b) + [D,E](a)b = a(D(E(b)) - E(D(b))) + (D(E(a))-E(D(a)))b$$

The result follows from rearranging terms. $\quad \square$

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(ii) Show that $D \circ E$ need not be a derivation.
Solution
By definition $D \circ E$ being a derivation means that for all $a, b \in A$ we have

$$(D \circ E) (ab) = a(D \circ E)(b) + (D \circ E)(a)b \\ \iff D(E(ab)) = aD(E(b)) + D(E(a))b \\ \iff D(aE(b) + E(a)b) = aD(E(b)) + D(E(a))b \\ \iff D(aE(b)) + D(E(a)b) = aD(E(b)) + D(E(a))b \\ \iff aD(E(b)) + D(a)E(b) + E(a)D(b) + D(E(a))b = aD(E(b)) + D(E(a))b \\ \iff D(a)E(b) + E(a)D(b) = 0$$

Let $D$ be ordinary differentiation in the associative algebra $A=C^{\infty}\mathbf{R}$ of infinitely differentiable functions $f,g: \mathbf{R} \to \mathbf{R}$, where the product $fg$ is given by pointwise multiplication: $(fg)(x) = f(x)g(x)$. Then $D \circ D$ is a derivation if and only if for every $a, b \in A$,

$$D(a)D(b) + D(a)D(b) = 0.$$

Take $a,b$ to be the identity functions that is $a(x) = x$, $b(x) = x$ for all $x \in \mathbf{R}$. Then

$$D(a)D(b) + D(a)D(b) = 1 \cdot 1 + 1 \cdot 1 = 2 \neq 0.$$

Indeed, we can check explicitly that

$$(D \circ D)(a^2) = D(D(a^2)) = D(2a) = 2.$$

If $D \circ D$ was a derivation we would have

$$(D \circ D)(aa) = a (D \circ D)(a) + (D \circ D)(a) a = 0,$$

and of course in $\mathbf{R}$ we have $0 \neq 2$ and this counterexample shows that the composition of two derivations is not necessarily a derivation. $\quad \square$

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Exercise 1.9
Let $L_1$ and $L_2$ be Lie algebras. Show that $L_1$ is isomorphic to $L_2$ if and only if there is a basis $B_1$ of $L_1$ and a basis $B_2$ of $L_2$ such that the structure constants of $L_1$ with respect to $B_1$ is equal to the structure constants of $L_2$ with respect to $B_2$.
Solution
$\left( \implies \right)$ Let $f: L_1 \to L_2$ be an isomorphism, and let $x_{1}, x_{2}, \ldots, x_{n}$ be a basis of $L_1$ (hold up- do we know that $L_1$ is finite dimensional?). (Check that you can do this) Then $f(x_1), \ldots, f(x_n)$ is a basis of $L_2$. $L_1$ has structure constants $a_{ij}^k$ such that

$$[x_i, x_j] = \sum_{n=0}^k a_{ij}^k x_{k} \\[4px] f([x_i, x_j]) = f\left(\sum_{n=0}^k a_{ij}^k x_{k}\right) \\[4px] [f(x_i), f(x_j)] = \sum_{n=0}^k f(a_{ij}^k x_k) = \sum_{n=0}^k a_{ij}^k f(x_k) \quad \square \\[4px]$$

$\left( \impliedby \right)$ Let $B_1 = (x_1, \ldots, x_n)$ be a basis of $L_1$ and let $B_2 = (y_1, \ldots, y_n)$ be a basis of $L_2$, with a shared set of structure constants $a_{ij}^k$. Let $f$ be the unique linear function satisfying

$$f(x_i) = y_i$$

Of course $f$ is an isomorphism of vector spaces, what we need to show is that $f$ is also an isomorphism of Lie algebras, that is that $f$ commutes with $[-,-]$, meaning that for all $a, b \in L_1$

$$f([a,b]) = [f(a), f(b)]$$

Use our basis to get

$$a = \lambda_1 x_1 + \cdots + \lambda_n x_n \\ b =\mu_1 x_1 + \cdots \mu_n x_n$$

Calculate that

$$f([a,b]) = f([\lambda_1x_1 + \cdots + \lambda_n x_n, \mu_1 x_1 + \cdots \mu_n x_n]) \\[4px] = f\left(\left[\lambda_1x_1, \sum_{i=1}^n \mu_i x_i \right] + \cdots + \left[\lambda_nx_n, \sum_{i=1}^n \mu_i x_i \right] \right) \\[8px] = f\left(\sum_{i=1}^n \left[ \lambda_i x_i, \sum_{j=1}^n \mu_j x_j \right] \right) = f\left(\sum_{i=1}^n \sum_{j=1}^n [\lambda_i x_i, \mu_jx_j]\right) \\[8px] = f\left(\sum_{i=1}^n \sum_{j=1}^n \lambda_i \mu_j [x_i, x_j]\right) = \sum_{i=1}^n \sum_{j=1}^n \lambda_i \mu_j f([x_i,x_j]) \\[8px]$$

Recall that by the definition of a structure constant

$$[x_i, x_j] = \sum_{k=1}^n a_{ij}^k x_k$$

So we get

$$f([a,b]) = \sum_{i=1}^n \sum_{j=1}^n \lambda_i \mu_j \sum_{k=1}^n a_{ij}^k x_k$$

Separately calculate that

$$[f(a),f(b)] = \left[f\left(\sum_{i=1}^n \lambda_i x_i \right),f\left(\sum_{j=1}^n \mu_j y_j\right)\right] = \left[ \sum_{i=1}^n \lambda_i f(x_i), \sum_{j=1}^n \mu_j f(x_j) \right] \\[8px] = \sum_{i=1}^n \sum_{j=1}^n [\lambda_i f(x_i), \mu_j f(x_j)] = \sum_{i=1}^n \sum_{j=1}^n \lambda_i \mu_j [f(x_i), f(x_j)]$$

By the definition of $f$ we get

$$[f(a), f(b)] = \sum_{i=1}^n \sum_{j=1}^n \lambda_i \mu_j [y_i, y_j]$$

By the hypothesis $B_1 = (x_1, \dots, x_n)$ and $B_2 = (y_1, \dots, y_n)$ share the same structure constants $a_{ij}^k$ so

$$[f(a), f(b)] = \sum_{i=1}^n \sum_{j=1}^n \lambda_i \mu_j \sum_{k=1}^n a_{ij}^k x_k$$

We conclude that for each $a, b \in L_1$ we have

$$f([a,b]) = [f(a), f(b)] \quad \square$$

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Exercise 1.10
Let $L$ be a Lie algebra with a basis $(x_1, \dots, x_n)$. What conditions does the Jacobi identity impose on the structure constants $a_{ij}^k$?
Solution
Recall that the Jacobi identity states that for all $x, y, z \in L$ we have

$$[x, [y, z]] + [y, [z, x]] + [z, [x, y]] = 0$$

and that, for each $i, j \in {1, \dots, n}$, the structure constants satisfy the equation

$$[x_i, x_j] = \sum_{k=1}^n a_{ij}^k x_k$$

Take $x = x_i$, $y = x_j$, $z = x_k$. Then

$$0 = [x,[y,z]] + [y,[z,x]] + [z, [x, y]] = [x_i, [x_j, x_k]] + [x_j, [x_k, x_i]] + [x_k, [x_i, x_j]] \\[8px] = \left[x_i, \sum_{l=1}^n a_{jk}^l x_l \right] + \left[x_j, \sum_{l=1}^n a_{ki}^l x_l \right] + \left[x_k, \sum_{l=1}^n a_{ij}^l x_l \right] \\ = \sum_{l=1}^n a_{jk}^l [x_i, x_l] + \sum_{l=1}^n a_{ki}^l [x_j, x_l ] + \sum_{l=1}^n a_{ij}^l[x_k, x_l ] \\ = \sum_{l=1}^n a_{ij}^l [x_i, x_l] + a_{ki}^l [x_j, x_l ] + a_{ij}^l[x_k, x_l] \\ = \sum_{l=1}^n a_{ij}^l \sum_{m=1}^n a_{il}^m x_m + a_{ki}^l \sum_{m=1}^n a_{jl}^mx_m + a_{ij}^l \sum_{m=1}^n a_{ik}^m x_m \\ = \sum_{l=1}^n \sum_{m=1}^n a_{ij}^l a_{il}^m x_m + a_{ki}^l a_{jl}^mx_m + a_{ij}^l a_{ik}^m x_m \\ = \sum_{m=1}^n \sum_{l=1}^n a_{ij}^l a_{il}^m x_m + a_{ki}^l a_{jl}^mx_m + a_{ij}^l a_{ik}^m x_m \\ = \sum_{m=1}^n \sum_{l=1}^n (a_{ij}^l a_{il}^m + a_{ki}^l a_{jl}^m + a_{ij}^l a_{ik}^m)x_m \\$$

Since $(x_1, \dots, x_n)$ forms a basis they are linearly independent, so for each $m \in 1, \dots, n$ we must have

$$\sum_{l=1}^n a_{ij}^l a_{il}^m + a_{ki}^l a_{jl}^m + a_{ij}^l a_{ik}^m = 0$$

---

Exercise 1.11
Let $L_1$ and $L_2$ be two abelian Lie algebras. Show that $L_1$ and $L_2$ are isomorphic if and only if they have the same dimension.
Solution
$(\implies)$ If $L_1$ and $L_2$ are isomorphic as Lie algebras than they must be isomorphic as vector spaces $\iff$ they have the same dimension
$(\impliedby)$ If $L_1$ and $L_2$ are the same dimension then we get a vector space isomorphism $f: L_1 \to L_2$. To show that this is a Lie algebra isomorphism, we need to show that for each $a, b \in L_1$ we have

$$f([a,b]) = [f(a),f(b)]$$

Since $L_1$ is abelian we have

$$f([a,b]) = f(0) = 0,$$

with the last equality coming from the fact that $f$ is a vector space isomorphism. $L_2$ is abelian as well, so

$$[f(a), f(b)] = 0.$$

We conclude that $f([a,b]) = [f(a),f(b)]$ for each $a, b \in L_1$, as desired. $\quad \square$

---

$1.12. \dag \quad$ Find the structure constants of $\textsf{sl}(2, F)$ with respect to the basis given by the matrices

$$e = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}, \> f = \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}, \> h = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}.$$

Solution

First calculate that

$$a_{ef}^e e + a_{ef}^f f + a_{ef}^h h = [e, f] = ef - fe \\[4px] = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix} - \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} \\[4px] = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} - \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} = 0e + 0f + 1h$$

Work similarly for the other combinations, avoiding duplicate work by recalling that $[a, b] = -[b,a]$. For $e$ and $h$ we get

$$a_{eh}^e e + a_{eh}^f f + a_{eh}^h h = [e, h] = eh - he \\[4px] = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} - \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} \\ = \begin{pmatrix} 0 & -1 \\ 0 & 0 \end{pmatrix} - \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} = -2e + 0f + 0h \\[4px]$$

For $f$ and $h$ we get

$$a_{fh}^e e + a_{fh}^f f + a_{fh}^h h = [f, h] = fh - hf \\[4px] = \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} - \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix} \\[4px] = \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix} - \begin{pmatrix} 0 & 0 \\ -1 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 2 & 0 \end{pmatrix} \\ = 0e + 2f + 0h.$$

---

$1.13 \quad$ Prove that $\textsf{sl}(2, \mathbf{C})$ has no non-trivial ideals.

Solution
Let $A$ be a non-zero ideal of $\sf{sl}(2, \bf{C})$. By definition, for each $a \in A$ and $b \in \sf{sl}(2, \bf{C})$ we have

$$[a, b] \in A$$

Using the basis from $1.12$,

$$a = a_ee + a_ff + a_hh, \\$$

for some $a_e, a_f, a_h \in \bf{C}$. Further, since we supposed that $A \neq 0$, we can take $a \neq 0$, so we do not have $a_e = a_f = a_h = 0$.

Calculate that

$$[a,e] = [a_ee + a_ff + a_hh, e] = a_e[e,e] + a_f[f,e] + a_h[h,e] \\ = -a_fh + 2a_h e \\ [a,f] = [a_ee + a_ff + a_hh, f] = a_e[e,f] + a_f[f,f] + a_h[h,f] \\ = a_e h - 2a_h f \\ [a, h] = [a_ee + a_ff + a_hh, h] = a_e[e,h] + a_f[f,h] + a_h[h,h] \\ = -2a_e e + 2 a_f f$$

Any linear combination of these must be in $A$, such as

$$a_f[a,f] + a_h[a,h] \\ = a_ea_fh - 2a_fa_hf - 2a_ea_he + 2a_fa_hf \\ = a_ea_fh - \cancel{2a_fa_hf} - 2a_ea_he + \cancel{2a_fa_hf} \\ = a_ea_fh - 2a_ea_he \\$$

Dividing by $a_e$ gives

$$a_fh - a_he \in A$$

Since

$$[a,e] = -a_fh + 2a_he \in A$$

Their sum is in the ideal as well,

$$a_h e \in A \iff a_h = 0 \text{ or } e \in A$$

If $a_h = 0$ then

$$[a,e] = -a_fh \in A \\ [a,f] = a_eh \in A$$

So

$$a_e = a_f = 0 \text{ or } h \in A$$

However $a_e = a_f = 0$ is a contradiction (since we already have $a_h = 0$ and explicitly can not have $a_e = a_f = a_h = 0$). Therefore we have $h \in A$.

Since $h \in A$ then $[e, h] = -2e \in A$ and $[f,h] = 2f \in A$. So we have $e, f, h \in A \implies A = \sf{sl}(2,\bf{C})$ since $\{e, f, h\}$ is a basis for $\sf{sl}(2,\bf{C})$.

We have shown that if $a_h = 0$ then $A = \sf{sl}(2, \bf{C})$. If $a_h \neq 0$ then $e \in A$, so $[e,f] = h \in A$ so we also have $A = \sf{sl}(2, \bf{C})$.

We have now shown that any non-zero ideal of $\sf{sl}(2,\bf{C})$ is equal to $\sf{sl}(2,\bf{C})$ itself. Therefore $\sf{sl}(2,\bf{C})$ has no non-trivial ideals. $\quad \square$

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$1.14. \dag \quad$ Let $L$ be the $3$-dimensional complex Lie algebra with basis $(x, y, z)$ and Lie bracket defined by

$$[x, y] = z, \> [y, z] = x, \> [z, x] = y.$$

(Here $L$ is the "complexification" of the $3$-dimensional real Lie algebra $\mathbf{R}^{3}_{\land})$

$(\text{i})$ Show that $L$ is isomorphic to the Lie subalgebra of $\textsf{gl}(3, \mathbf{C})$ consisting of all $3 \times 3$ antisymmetric matrices with entries in $\mathbf{C}$.

$(\text{ii})$ Find an explicit isomorphism $\textsf{sl}(2, \mathbf{C}) \cong L$.

Solution

To solve $(\text{i})$ we can make use of Exercise $1.9$, which tells us that it is sufficient to find a basis of the $3 \times 3$ antisymmetric matrices with entries in $\bf{C}$ $\{x, y, z\}$ such that

$$[x,y] = z, \> [y,z] = x, \> [z,x] = y$$

Let

$$x = \begin{pmatrix} 0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}, \> y = \begin{pmatrix} 0 & 0 & -1 \\ 0 & 0 & 0 \\ 1 & 0 & 0 \end{pmatrix}, \> z = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & -1 & 0 \end{pmatrix},$$

then $\{x, y, z\}$ is a basis for the $3 \times 3$ antisymmetric matrices because they are clearly linearly independent and

$$\begin{pmatrix} 0 & a & -b \\ -a & 0 & c \\ b & -c & 0 \end{pmatrix} = ax + by + cz,$$

so it is sufficient to show that

$$[x,y] = z, \> [y,z] = x, \> [z,x] = y.$$

Calculate that

$$[x,y] = xy - yx \\[4px] = \begin{pmatrix} 0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} \begin{pmatrix} 0 & 0 & -1 \\ 0 & 0 & 0 \\ 1 & 0 & 0 \end{pmatrix} -\begin{pmatrix} 0 & 0 & -1 \\ 0 & 0 & 0 \\ 1 & 0 & 0 \end{pmatrix} \begin{pmatrix} 0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} \\[4px] = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0\\ \end{pmatrix} - \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & -1 & 0 \end{pmatrix} = z, \\[4px] [y,z] = yz - zy \\[4px] = \begin{pmatrix} 0 & 0 & -1 \\ 0 & 0 & 0 \\ 1 & 0 & 0 \end{pmatrix} \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & -1 & 0 \end{pmatrix} - \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & -1 & 0 \end{pmatrix} \begin{pmatrix} 0 & 0 & -1 \\ 0 & 0 & 0 \\ 1 & 0 & 0 \end{pmatrix} \\[4px] = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end{pmatrix} - \begin{pmatrix} 0 & 0 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} = x, \\[4px] [z,x] = zx - xz \\[4px] \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & -1 & 0 \end{pmatrix} \begin{pmatrix} 0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} - \begin{pmatrix} 0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & -1 & 0 \end{pmatrix} \\[4px] = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 1 & 0 & 0 \end{pmatrix} - \begin{pmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 0 & -1 \\ 0 & 0 & 0 \\ 1 & 0 & 0 \end{pmatrix} = y.$$

This completes $\text{(i)}$. Recall that $\text{(ii)}$ asks for an explicit isomorphism $\varphi: \sf{sl}(2,\bf{C}) \to \mathnormal{L}$. In other words, $\varphi$ has to be a linear bijective map that respects the Lie bracket, so for each $u, v \in \sf{sl}(2, \bf{C})$

$$[\varphi(u), \varphi(v)] = \varphi([u, v])$$

Recall from $1.12$ that

$$\left\{ e = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}, \> f = \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}, \> h = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \right\}$$

is a basis for $\sf{sl}(2, \bf{C})$ and that

$$[e,f] = h, \> [e, h] = -2e, \> [f,h] = 2f,$$

so it is sufficient to show that (prove this?)

$$[\varphi(e), \varphi(f)] = \varphi([e, f]) = \varphi(h), \\ [\varphi(e), \varphi(h)] = \varphi([e,h]) = -2 \varphi(e), \\ [\varphi(f), \varphi(h)] = \varphi([f,h]) = 2 \varphi(f).$$

The key insight comes from the following calculation

$$[ix + z, ix - z] = [ix, ix - z] + [z, ix - z] \\ = [ix, ix] - [ix, z] + [z, ix] - [z,z] \\ = -2i [x,z] = 2iy$$

So take (prove that this is well defined and a bijection? Enough to show that these are three linearly independent elements?)

$$\varphi(e) = ix + z, \> \varphi(f) = ix - z, \> \varphi(h) = 2iy.$$

The previous calculation verified that

$$[\varphi(e), \varphi(f)] = \varphi(h),$$

so one equation down, two to go. Calculate that

$$[\varphi(e), \varphi(h)] = [ix + z, 2iy] = [ix, 2iy] + [z, 2iy] \\ = -2[x,y] + 2i [z,y] = -2ix - 2z = -2 \varphi(e).$$

One more to go!

$$[\varphi(f), \varphi(h)] = [ix - z, 2iy] = [ix, 2iy] - [z, 2iy] \\ = -2[x,y] - 2i [z,y] = 2ix - 2z = 2(ix - z) = 2 \varphi(f). \quad \square$$

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$1.15. \quad$ Let $S$ be an $n \times n$ matrix with entries in a field $F$. Define

$$\mathsf{gl}_S (n, F) = \{ x \in \mathsf{gl}(n, F): x^tS = -Sx \}.$$

$(\text{i}) \>$ Show that $\mathsf{gl}_S(n, F)$ is a Lie subalgebra of $\mathsf{gl}(n, F)$.
$(\text{ii}) \>$ Find $\mathsf{gl}_S(2, \bf{R})$ if $S = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$.
$(\text{iii}) \>$ Does there exist a matrix $S$ such that $\mathsf{gl}_S(2,\bf{R})$ is equal to the set of all diagonal matrices in $\mathsf{gl}(2,\bf{R})$?
$(\text{iv}) \>$ Find a matrix $S$ such that $\mathsf{gl}_S (3,\bf{R})$ is isomorphic to the Lie algebra $\mathbf{R}^3_\land$ defined in $\S 1.2$, Example 1.
Hint: Part $(\text{i})$ of Exercise $1.14$ is relevant.

Solution

$\text{(i)} \>$ In order to show that $\mathsf{gl}_S(n,F)$ is a Lie subalgebra of $\mathsf{gl}(n,F)$, we must first show that it is a vector subspace.
Let $\lambda \in F$ and $a, b \in \mathsf{gl}_S(n,F)$. It suffices to show that $a + \lambda b \in \mathsf{gl}_S(n,F)$.
By the definition of $\mathsf{gl}_S(n,F)$ we have

$$\tag{1} a^tS = -Sa, \> b^tS = -Sb$$

Since matrix transposition is linear

$$(a + \lambda b)^t = a^t + \lambda b^t$$

Multiplying by $S$

$$(a + \lambda b)^tS = (a^t + \lambda b^t)S = a^tS + \lambda b^t S$$

Plugging in $(1)$

$$= -Sa - \lambda Sb = -S(a + \lambda b)$$

This shows that $\mathsf{gl}_S(n,F)$ is a vector subspace of $\mathsf{gl}(n,F)$. To show that it is a Lie subalgebra it remains to show that $[a,b] \in \mathsf{gl}_S(n,F)$. Calculate that

$$[a,b]^t S = (ab - ba)^t S = (ab)^tS - (ba)^tS \\ = b^ta^tS - a^tb^tS = b^t (-Sa) - a^t (-Sb) \\ = (a^t S)b - (b^tS)a = -Sab - (-Sb)a \\ = Sba - Sab = S (ba - ab) = -S(ab - ba) = -S [a,b]. \quad \square$$

$\text{(ii)} \>$ Let $S = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$, we need to find $\mathsf{gl}_S(2,\mathbf{R})$. Let $x = \begin{pmatrix} x_{11} & x_{12} \\ x_{21} & x_{22} \end{pmatrix}$. Calculate that

$$x^tS = -Sx \\[4px] \iff \begin{pmatrix} x_{11} & x_{21} \\ x_{12} & x_{22}\end{pmatrix} \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} = - \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} \begin{pmatrix} x_{11} & x_{12} \\ x_{21} & x_{22} \end{pmatrix} \\[4px] \iff \begin{pmatrix} 0 & x_{11} \\ 0 & x_{12} \end{pmatrix} = - \begin{pmatrix} x_{21} & x_{22} \\ 0 & 0 \end{pmatrix} \\[4px] \iff \begin{pmatrix} x_{21} & x_{11} + x_{22} \\ 0 & x_{12} \end{pmatrix} = 0$$

So we can conclude that $\mathsf{gl}_S(2,\mathbf{R}) = \text{span}\left\{\begin{pmatrix}1 & 0 \\ 0 & -1\end{pmatrix}\right\}$. $\quad \square$

$\text{(iii)} \>$ We need to find or prove that their doesn't exist a matrix $S$ such that $\mathsf{gl}_S(2,\mathbf{R}) = \text{span}\left(\left\{\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix},\begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}\right\}\right)$.

$$x^tS = -Sx \\ \begin{pmatrix} x_{11} & x_{12} \\ x_{21} & x_{22} \end{pmatrix}^\intercal \begin{pmatrix} S_{11} & S_{12} \\ S_{21} & S_{22} \end{pmatrix} = - \begin{pmatrix} S_{11} & S_{12} \\ S_{21} & S_{22} \end{pmatrix} \begin{pmatrix} x_{11} & x_{12} \\ x_{21} & x_{22} \\ \end{pmatrix} \\[4px] \begin{pmatrix} x_{11} & x_{21} \\ x_{12} & x_{22} \end{pmatrix} \begin{pmatrix} S_{11} & S_{12} \\ S_{21} & S_{22} \end{pmatrix} = - \begin{pmatrix} S_{11} & S_{12} \\ S_{21} & S_{22} \end{pmatrix} \begin{pmatrix} x_{11} & x_{12} \\ x_{21} & x_{22} \\ \end{pmatrix} \\[4px] \begin{pmatrix} x_{11}S_{11} + x_{21}S_{21} & x_{11}S_{12} + x_{21}S_{22} \\ x_{12}S_{11} + x_{22}S_{21} & x_{12}S_{12} + x_{22}S_{22} \end{pmatrix} \\ = - \begin{pmatrix} S_{11}x_{11} + S_{12}x_{21} & S_{11}x_{12} + S_{12}x_{22} \\ S_{21}x_{11} + S_{22}x_{21} & S_{21}x_{12} + S_{22}x_{22} \end{pmatrix} \\ x_{11}S_{11} + x_{21}S_{21} + S_{11}x_{11} + S_{12}x_{21} = 0 \\ x_{11}S_{12} + x_{21}S_{22} + S_{11}x_{12} + S_{12}x_{22} = 0 \\ x_{12}S_{11} + x_{22}S_{21} + S_{21}x_{11} + S_{22}x_{21} = 0 \\ x_{12}S_{12} + x_{22}S_{22} + S_{21}x_{12} + S_{22}x_{22} = 0$$

We need to choose $S$ such that this reduces to

$$x_{12} = x_{21} = 0$$

The last equation can be rewritten as

$$(S_{12} + S_{21})x_{12} + 2S_{22}x_{22} = 0$$

And the first as

$$(S_{12} + S_{21})x_{21} + 2S_{11}x_{11} = 0$$

So we must have

$$S_{11} = S_{22} = 0$$

Plugging in

$$x_{21}(S_{21} + S_{12}) = 0 \\ x_{11}S_{12} + x_{22}S_{12} = 0 \\ x_{22}S_{21} + x_{11}S_{21} = 0 \\ x_{12}(S_{12} + S_{21}) = 0$$

so we would need to also have $S_{12} = S_{21}$, but this doesn't give us the desired result, so no such matrix $S$ exists. $\quad \square$

$\text{(iv)} \quad$ We need to find $S$ such that $\mathsf{gl}_S(3,\mathbf{R})$ is isomorphic to $\mathbf{R}^3_\land$. By part $\text{(i)}$ of Exercise $1.14$ we know that $\mathbf{R}^3_\land$ is isomorphic to the antisymmetric $3 \times 3$ matrices. So we need to find $S$ such that

$$x^tS = -Sx \iff x^t = -x$$

Of course $S = I$ solves this!

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$1.16. \dag \quad$ Show, by giving an example, that if $F$ is a field of characteristic $2$, there are algebras over $F$ which satisfy $(\text{L1'})$ and $(\text{L}2)$ but are not Lie algebras.

Solution

Recall that

$$\tag{L1'} \phantom{(L1')} \quad [x,y] = -[y,x] \quad \text{for all} \enspace x,y \in L.$$

$$\tag{L2} \phantom{(L2)} \quad [x,[y,z]] + [y,[z,x]] + [z,[x,y]] = 0 \quad \text{for all} \enspace x,y,z \in L.$$

So let's find and an algebra over $F$ such that the above holds but $\text{(L1)}$

$$\tag{L1} \phantom{(L1')} \quad [x,x] = 0 \quad \text{for all} \enspace x \in L$$

does not.

Consider the vector space $V$ of $3 \times 3$ diagonal matrices with entries in $F$ and over the field of $F$. Its basis is given by

$$e = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} , \enspace j = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{pmatrix}, \enspace k = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix}, \\[.35em] V = \text{span} \left( \left\{ e,j,k \right\} \right).$$

Equip it with a bilinear map given by

$$[x,y] = xy.$$

Calculate that

$$[e,e] = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} = e,$$

$$[e,j] = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} \begin{pmatrix} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} = 0,$$

$$[e,k] = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix} = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} = 0,$$

$$[j,e] = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{pmatrix} \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} = 0,$$

$$[j,j] = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{pmatrix} \begin{pmatrix} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{pmatrix} = j,$$

$$[j,k] = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{pmatrix} \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix} = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} = 0,$$

$$[k,e] = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} = 0,$$

$$[k,j] = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} = 0,$$

$$[k,k] = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix} = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix} = k.$$

Obviously $\text{(L1)}$ does not hold, so it remains to show that $\text{(L1')}$ and $\text{(L2)}$ hold.

Let $x, y \in V$, and let $x = \lambda_e e + \lambda_j j + \lambda_k k, \> y = \mu_e e + \mu_j j + \mu_k k$. Calculate that

$$\begin{aligned} &\phantom{=} [x,y] = [\lambda_e e + \lambda_j j + \lambda_k k, \mu_e e + \mu_j j + \mu_k k] \\ &= \lambda_e \mu_e [e,e] + \lambda_e \mu_j[e,j] + \lambda_e \mu_k [e,k] \\ &\phantom{=} + \enspace \lambda_j \mu_e [j,e] + \lambda_j \mu_j [j,j] + \lambda_j \mu_k [j,k] \\ &\phantom{=} + \enspace \lambda_k \mu_e [k,e] + \lambda_k \mu_j [k,j] + \lambda_k \mu_k [k,k] \\ &= \lambda_e \mu_e e + \lambda_j \mu_j j + \lambda_k \mu_k k. \end{aligned}$$

By the symmetry of the above equation, $[x,y] = [y,x]$, so

$$[x,y] + [y,x] = [x,y] + [x,y] = (1 + 1) [x,y] = 0,$$

therefore $\text{(L1')}$ holds. Calculate that

$$\begin{aligned} &\phantom{=}[e,[j,k]] + [j, [k,e]] + [k, [e,j]] \\ &= [e,0] + [j,0] + [k,0] \\ &= 0 + 0 + 0 = 0, \end{aligned}$$

so $\text{(L2)}$ holds as well. $\enspace \square$

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$1.17. \quad$ Let $V$ be an $n$-dimensional complex vector space and let $L = \mathsf{gl}(V)$. Suppose that $x \in L$ is diagonalisable, with eigenvalues $\lambda_1, \dots, \lambda_n$. Show that $\text{ad } x \in \mathsf{gl}(L)$ is also diagonalisable, and that its eigenvalues are $\lambda_i - \lambda_j$ for $1 \leq i, j \leq n$.

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$1.18. \quad$ Let $L$ be a Lie algebra. We saw in $\S 1.6$, Example $1.2(2)$ that the maps $\text{ad } x: L \to L$ for $x \in L$ are derivations of $L$; these are known as inner derivations. Show that if $\text{IDer } L$ is the set of inner derivations of $L$, then $\text{IDer } L$ is an ideal of $\text{Der } L$.

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$1.19. \quad$ Let $A$ be an algebra and let $\delta: A \to A$ be a derivation. Prove that $\delta$ satisfies the Leibniz rule

$$\delta^n(xy) = \sum_{r = 0}^n \binom{n}{r} \delta^r(x) \delta^{n-r}(y) \quad \text{for all } x,y \in A$$

$$\\$$