Physlib.Mathematics.Groups.SpecialUnitary.LieAlgebra.Adjoint

For a natural number $n$, the adjoint representation of the special unitary group $SU(n)$ on its real Lie algebra $\mathfrak{su}(n)$ is the group representation where an element $g \in SU(n)$ acts on $x \in \mathfrak{su}(n)$ via conjugation. Specifically, the Lie algebra $\mathfrak{su}(n)$ is the space of $n \times n$ complex matrices that are traceless and hermitian, and the action is given by $x \mapsto g x g^\dagger$, where $g^\dagger$ denotes the conjugate transpose of $g$.

For any natural number $n$, let $g$ be an element of the special unitary group $SU(n)$ and $x$ be an element of the Lie algebra $\mathfrak{su}(n)$ (consisting of $n \times n$ traceless hermitian matrices). The matrix corresponding to the adjoint action of $g$ on $x$ is given by $g x g^\dagger$, where $g^\dagger$ denotes the conjugate transpose of $g$.

For any element $g$ of the special unitary group $SU(n)$ and any $x, y$ in the Lie algebra $\mathfrak{su}(n)$, the adjoint representation $Ad_g$ preserves the Lie bracket $[\cdot, \cdot]$, satisfying the identity: $$Ad_g([x, y]) = [Ad_g(x), Ad_g(y)]$$ where $Ad_g(x) = g x g^\dagger$ is the action by conjugation and the Lie bracket on the space of traceless hermitian matrices is defined as $[x, y] = i(xy - yx)$.

For any natural number $n$ and any element $g$ of the special unitary group $SU(n)$, the adjoint representation of $g$ on the Lie algebra $\mathfrak{su}(n)$ is the identity operator if and only if there exists a complex scalar $c \in \mathbb{C}$ such that the matrix representation of $g$ is $c I_n$, where $I_n$ is the $n \times n$ identity matrix. In other words, the kernel of the adjoint representation $\text{Ad} : SU(n) \to \text{Aut}(\mathfrak{su}(n))$ consists exactly of the scalar matrices in $SU(n)$.

Let $SU(n)$ be the special unitary group and $\mathfrak{su}(n)$ be its Lie algebra of traceless hermitian matrices. For any two elements $g, h \in SU(n)$, their adjoint actions on $\mathfrak{su}(n)$ are equal if and only if there exists a complex scalar $c \in \mathbb{C}$ such that $g = c h$ as matrices. This implies that the adjoint representation is faithful on $SU(n)$ modulo its center.

For a natural number $n$, the complexified adjoint representation of the special unitary group $SU(n)$ is a representation over the field $\mathbb{C}$ acting on the complexified Lie algebra $\mathbb{C} \otimes_{\mathbb{R}} \mathfrak{su}(n)$. It is defined by extending the real adjoint representation on $\mathfrak{su}(n)$ (where an element $g \in SU(n)$ acts on $x \in \mathfrak{su}(n)$ via conjugation $g x g^\dagger$) to its complexification via base change. Specifically, for $g \in SU(n)$, its action on a tensor $z \otimes x \in \mathbb{C} \otimes_{\mathbb{R}} \mathfrak{su}(n)$ is given by $z \otimes (g x g^\dagger)$.

Let $n$ be a natural number, $g \in SU(n)$ be an element of the special unitary group, $z \in \mathbb{C}$ be a complex scalar, and $x \in \mathfrak{su}(n)$ be an element of the Lie algebra of traceless hermitian matrices. The action of the complexified adjoint representation $\text{Ad}_g^{\mathbb{C}}$ on the pure tensor $z \otimes x$ in the complexified Lie algebra $\mathbb{C} \otimes_{\mathbb{R}} \mathfrak{su}(n)$ is given by $$\text{Ad}_g^{\mathbb{C}}(z \otimes x) = z \otimes \text{Ad}_g(x),$$ where $\text{Ad}_g$ denotes the adjoint representation of $SU(n)$ on $\mathfrak{su}(n)$.

For any element $g$ of the special unitary group $SU(n)$ and any elements $A, B$ in the complexified Lie algebra $\mathbb{C} \otimes_{\mathbb{R}} \mathfrak{su}(n)$, the complexified adjoint representation $\text{Ad}_g$ preserves the Lie bracket, such that: $$\text{Ad}_g([A, B]) = [\text{Ad}_g(A), \text{Ad}_g(B)]$$ where $[\cdot, \cdot]$ denotes the Lie bracket on the complexified Lie algebra.

Let $n$ be a natural number and $g$ be an element of the special unitary group $SU(n)$. The complexified adjoint representation $\text{Ad}_{\mathbb{C}}(g)$ is the identity map if and only if $g$ is a scalar matrix, meaning there exists a complex number $c \in \mathbb{C}$ such that $g = c I_n$, where $I_n$ is the $n \times n$ identity matrix.

For any two elements $g, h$ of the special unitary group $SU(n)$, their complexified adjoint actions are equal if and only if there exists a complex scalar $c \in \mathbb{C}$ such that $g = ch$ as matrices. Here, the complexified adjoint representation acts on the complexified Lie algebra $\mathbb{C} \otimes_{\mathbb{R}} \mathfrak{su}(n)$ via the extension of the conjugation map $x \mapsto g x g^\dagger$.

Let $g$ be an element of the special unitary group $SU(n)$ and $A$ be an element of the complexified Lie algebra $\mathbb{C} \otimes_{\mathbb{R}} \mathfrak{su}(n)$. The matrix representation of the complexified adjoint action of $g$ on $A$ is given by the conjugation $g M g^\dagger$, where $M$ is the matrix corresponding to $A$ and $g^\dagger$ is the conjugate transpose of $g$.

Let $g \in SU(n)$ be a diagonal matrix with diagonal entries $d_i \in \mathbb{C}$ for $i \in \{1, \dots, n\}$. For any element $A$ in the complexified Lie algebra $\mathfrak{su}(n)_{\mathbb{C}}$, the $(x, y)$-entry of the matrix resulting from the adjoint action of $g$ on $A$ is given by $$(g \cdot A)_{xy} = d_x \overline{d_y} A_{xy}$$ where $A_{xy}$ denotes the $(x, y)$-entry of $A$ and $\overline{d_y}$ is the complex conjugate of $d_y$.

Let $SU(n)$ be the special unitary group and $\mathfrak{su}(n)_{\mathbb{C}}$ be the complexification of the Lie algebra of traceless hermitian matrices. For any $g \in SU(n)$ and any element $A \in \mathfrak{su}(n)_{\mathbb{C}}$, the trace of the matrix representing the adjoint action of $g$ on $A$ is equal to the trace of the matrix representing $A$: $$\text{Tr}(\text{Ad}_g A) = \text{Tr}(A)$$ where $\text{Ad}_g A$ is defined by the conjugation $gAg^{\dagger}$.

Let $\mathfrak{su}(n)$ be the Lie algebra of $n \times n$ traceless hermitian matrices over $\mathbb{C}$. The map `adjointContr` is the invariant bilinear form defined as an intertwining map from the tensor product of the adjoint representation of $SU(n)$ with itself to the trivial representation $\mathbb{R}$. For elements $x, y \in \mathfrak{su}(n)$, the contraction is given by the real part of the trace of their product, i.e., $x \otimes y \mapsto \text{Re}(\text{Tr}(xy))$. Since $x$ and $y$ are hermitian matrices, $\text{Tr}(xy)$ is always real, and the map corresponds to the standard trace form on the Lie algebra.

Let $\mathfrak{su}(n)$ be the Lie algebra of $n \times n$ complex matrices that are traceless and hermitian. For any two elements $x, y \in \mathfrak{su}(n)$, the invariant bilinear form $\text{adjointContr}$ evaluated on their tensor product $x \otimes y$ is equal to the real part of the trace of the matrix product $xy$: $$\text{adjointContr}(x \otimes y) = \text{Re}(\text{Tr}(xy))$$

For any natural number $n$, let $g \in SU(n)$ be an element of the special unitary group and $x, y \in \mathfrak{su}(n)$ be elements of its Lie algebra of traceless hermitian matrices. The adjoint representation $\text{Ad}_g$ preserves the real contraction (invariant trace form) $\text{adjointContr}$, such that: $$\text{adjointContr}(\text{Ad}_g x \otimes \text{Ad}_g y) = \text{adjointContr}(x \otimes y)$$ where the adjoint action is given by conjugation $x \mapsto g x g^\dagger$.

For any natural number $n$ and any two elements $x, y$ in the Lie algebra $\mathfrak{su}(n)$ (the space of $n \times n$ traceless hermitian matrices), the complex trace of the product of their underlying matrices, $\text{Tr}(xy)$, is equal to their real-valued contraction $\text{adjointContr}(x, y)$. This reflects the fact that the trace of the product of two hermitian matrices is always real.

Let $\mathfrak{su}(n)$ be the Lie algebra of $n \times n$ complex matrices that are traceless and hermitian. For any $x, y \in \mathfrak{su}(n)$, the adjoint contraction map $\text{adjointContr}$ (the invariant trace form $x \otimes y \mapsto \text{Tr}(xy)$) is symmetric, such that $\text{adjointContr}(x \otimes y) = \text{adjointContr}(y \otimes x)$.

Let $\mathfrak{su}(n)$ be the Lie algebra of $n \times n$ traceless hermitian matrices. For any $x, y, z \in \mathfrak{su}(n)$, the real invariant trace form `adjointContr` (the bilinear form defined by the trace of the product of matrices) is invariant under the Lie bracket $\lfloor \cdot, \cdot \rfloor$, satisfying: $$\text{adjointContr}(\lfloor x, y \rfloor \otimes z) = \text{adjointContr}(x \otimes \lfloor y, z \rfloor)$$ In terms of the matrix trace, this is equivalent to $\text{Tr}(\lfloor x, y \rfloor z) = \text{Tr}(x \lfloor y, z \rfloor)$.

For any natural number $n$ and any $x$ in the Lie algebra $\mathfrak{su}(n)$ (the space of $n \times n$ traceless hermitian complex matrices), the invariant trace form evaluated at $x \otimes x$ is non-negative, i.e., $0 \le \text{Tr}(x^2)$.

Let $\mathfrak{su}(n)$ be the Lie algebra of $n \times n$ complex matrices that are traceless and hermitian. For any $x \in \mathfrak{su}(n)$, the contraction (trace form) $\text{Tr}(x^2)$ is equal to zero if and only if $x$ is the zero matrix.

Let $\mathfrak{su}(n)$ be the Lie algebra of $n \times n$ complex matrices that are traceless and hermitian. For any non-zero matrix $x \in \mathfrak{su}(n)$, the invariant trace form evaluated at $x \otimes x$, given by $\text{Tr}(x^2)$, is strictly positive.

For any natural number $n$, let $\mathfrak{su}(n)$ be the Lie algebra of $n \times n$ complex matrices that are traceless and hermitian. For any $x \in \mathfrak{su}(n)$, if the trace form $\text{Tr}(xy) = 0$ for all $y \in \mathfrak{su}(n)$, then $x = 0$.

Let $\mathfrak{su}(n)$ be the Lie algebra of $n \times n$ complex matrices that are traceless and hermitian. Let $\langle x, y \rangle = \text{Tr}(xy)$ denote the invariant trace form on $\mathfrak{su}(n)$. If an element $y \in \mathfrak{su}(n)$ satisfies $\text{Tr}(xy) = 0$ for all $x \in \mathfrak{su}(n)$, then $y = 0$.

For any natural number $n$, let $\mathfrak{su}(n)$ be the Lie algebra of $n \times n$ complex matrices that are traceless and hermitian. The linear map that sends each matrix $y \in \mathfrak{su}(n)$ to the linear functional $f_y : \mathfrak{su}(n) \to \mathbb{R}$ defined by $f_y(x) = \text{Tr}(xy)$ is injective.

For any natural number $n$, let $\mathfrak{su}(n)$ be the Lie algebra of $n \times n$ traceless hermitian matrices. The real contraction `adjointContr`, viewed as a bilinear form $B: \mathfrak{su}(n) \times \mathfrak{su}(n) \to \mathbb{R}$ defined by $B(x, y) = \text{Tr}(xy)$, is symmetric. That is, for all $x, y \in \mathfrak{su}(n)$, $B(x, y) = B(y, x)$.

For any natural number $n$, the invariant bilinear form on the Lie algebra $\mathfrak{su}(n)$ (consisting of $n \times n$ traceless hermitian matrices), defined by $B(x, y) = \text{Tr}(xy)$, is nondegenerate.

The map `adjointℂContr` is an intertwining map from the tensor product of the complexified adjoint representation of $SU(n)$ with itself to the trivial representation over $\mathbb{C}$. It is the complexification of the real invariant trace form `adjointContr`. For elements $A, B$ in the complexified Lie algebra $\mathfrak{su}(n)_{\mathbb{C}} = \mathbb{C} \otimes_{\mathbb{R}} \mathfrak{su}(n)$, the contraction is given by the trace of their product, $A \otimes B \mapsto \text{Tr}(AB)$.

For any element $g$ in the special unitary group $SU(n)$ and any elements $A, B$ in the complexified Lie algebra $\mathfrak{su}(n)_{\mathbb{C}}$, the complexified contraction (trace form) is invariant under the complexified adjoint representation $\text{Ad}_g$: $$\text{Tr}(\text{Ad}_g A \otimes \text{Ad}_g B) = \text{Tr}(A \otimes B)$$ where $\text{Tr}(A \otimes B)$ denotes the trace of the product of the matrices corresponding to $A$ and $B$.

For any elements $A$ and $B$ in the complexified Lie algebra $\mathfrak{su}(n)_{\mathbb{C}}$, the contraction $\text{adjointℂContr}$ applied to their tensor product $A \otimes B$ is equal to the trace of the product of the matrices representing $A$ and $B$. That is, $\text{adjointℂContr}(A \otimes B) = \text{Tr}(AB)$.

For any natural number $n$, let $\mathfrak{su}(n)$ be the Lie algebra of $n \times n$ traceless hermitian matrices. For any $x, y \in \mathfrak{su}(n)$, the complexified invariant trace form $\text{adjointℂContr}$ evaluated on the tensors $1 \otimes x$ and $1 \otimes y$ in the complexified Lie algebra $\mathbb{C} \otimes_{\mathbb{R}} \mathfrak{su}(n)$ is equal to the real invariant trace form $\text{adjointContr}$ evaluated on $x \otimes y$: $$\text{adjointℂContr}((1 \otimes x) \otimes (1 \otimes y)) = \text{adjointContr}(x \otimes y)$$

Let $\mathfrak{su}(n)_{\mathbb{C}}$ be the complexification of the Lie algebra $\mathfrak{su}(n)$, defined as $\mathbb{C} \otimes_{\mathbb{R}} \mathfrak{su}(n)$. For any two elements $A, B \in \mathfrak{su}(n)_{\mathbb{C}}$, the complexified adjoint contraction map `adjointℂContr` (which corresponds to the trace form $\text{Tr}(AB)$) is symmetric: $$\text{adjointℂContr}(A \otimes B) = \text{adjointℂContr}(B \otimes A)$$

Let $n$ be a natural number and $\mathfrak{su}(n)_{\mathbb{C}}$ be the complexification of the Lie algebra $\mathfrak{su}(n)$. For any elements $A, B, C \in \mathfrak{su}(n)_{\mathbb{C}}$, the complexified trace form satisfies the invariance property under the Lie bracket: $$\text{Tr}([A, B] C) = \text{Tr}(A [B, C])$$ where $[ \cdot, \cdot ]$ denotes the Lie bracket on the complexified algebra.

For any natural number $n$, let $\mathfrak{su}(n)_{\mathbb{C}}$ denote the complexification of the Lie algebra $\mathfrak{su}(n)$. For an element $A \in \mathfrak{su}(n)_{\mathbb{C}}$, if the complexified invariant trace form $\text{Tr}(AB) = 0$ for all $B \in \mathfrak{su}(n)_{\mathbb{C}}$, then $A = 0$.

Let $\mathfrak{su}(n)_{\mathbb{C}} = \mathbb{C} \otimes_{\mathbb{R}} \mathfrak{su}(n)$ be the complexification of the Lie algebra of $SU(n)$, consisting of $n \times n$ complex matrices that are the complexification of traceless hermitian matrices. Let the complexified trace form (contraction) be denoted by $\langle A, B \rangle = \text{Tr}(AB)$. For any element $B \in \mathfrak{su}(n)_{\mathbb{C}}$, if $\langle A, B \rangle = 0$ for all $A \in \mathfrak{su}(n)_{\mathbb{C}}$, then $B = 0$.

In the complexified Lie algebra $\mathfrak{su}(n)_{\mathbb{C}}$, the linear map that sends an element $B \in \mathfrak{su}(n)_{\mathbb{C}}$ to the linear functional $A \mapsto \text{Tr}(AB)$ (defined by the complexified trace form) is injective.

Let $n$ be a natural number and let $\mathfrak{su}(n)_{\mathbb{C}} = \mathbb{C} \otimes_{\mathbb{R}} \mathfrak{su}(n)$ be the complexification of the Lie algebra of $n \times n$ traceless hermitian matrices. The bilinear form $B: \mathfrak{su}(n)_{\mathbb{C}} \times \mathfrak{su}(n)_{\mathbb{C}} \to \mathbb{C}$ induced by the complex contraction `adjointℂContr` (which maps $A \otimes B \mapsto \text{Tr}(AB)$) is symmetric. That is, for all $A, B \in \mathfrak{su}(n)_{\mathbb{C}}$, $B(A, B) = B(B, A)$.

Let $n$ be a natural number and $\mathfrak{su}(n)_{\mathbb{C}} = \mathbb{C} \otimes_{\mathbb{R}} \mathfrak{su}(n)$ be the complexification of the Lie algebra of $n \times n$ traceless hermitian matrices. The complex bilinear form $B: \mathfrak{su}(n)_{\mathbb{C}} \times \mathfrak{su}(n)_{\mathbb{C}} \to \mathbb{C}$ defined by the trace of the product of two matrices, $B(A, B) = \text{Tr}(AB)$, is nondegenerate.