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Cosmas Zachos
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I don't have that text, but your question is standard stuff. I suspect what you are missing is the fast interplay of D-dimensional vectors and the D×D matrices acting on them, sometimes regarded as vector elements themselves, of a larger vector!

The dimension D of the representation is the dimensionality of the "vectors" on which the D×D representation matrices act on. So,But all kinds of objects "furnish" D-dimensional vectors on which the suitable representation matrices act to transform them. "Furnish" means provide components of a large D-dimensional vector specifying a possibly new representation.

Antisymmetric tensors $T^{ij}$ in $SO(N)$ with indices $j=1,...,N$ comprise a set of $\frac{1}{2}N(N-1)$ independent antisymmetric matrices, so they furnish vectors of a an irreducible D-dimensional representation, with $D=\frac{1}{2}N(N-1)$: they define the relevant vector space. Each such matrix is a component of the D-vector.

It is also true that there are $\frac{1}{2}N(N-1)$ generators of $SO(N)$, indexed by the (mn) pairs, $$\mathcal{J}_{(m n)}^{i j}=\left(\delta^{m i} \delta^{n j}-\delta^{m j} \delta^{n i}\right)$$ which act on the space of N-dimensional js to rotate them to such is.

  • However, these N×N antisymmetric matrices may also be regarded as the the $D=\frac{1}{2}N(N-1)$ components of a new vector, acted upon by some larger $\frac{1}{2}N(N-1) \times \frac{1}{2}N(N-1)$ matrices, ( actuallyactually $\frac{1}{2}N(N-1)$ independent such, indexed by (mn). This is the adjoint representation.)

So the D=N generators became the $D=\frac{1}{2}N(N-1)$-vector elements of the adjoint (which they furnished). Fortuitously, these two dimensions are the same for N=3, so the fundamental of SO(3) is happens to also thebe its adjoint!

Now

We can also regard the antisymmetric tensor $T^{ij}$ as an $N$-by-$N$ matrix, and hence write it as a linear combination of the $\mathcal{J}_a$'s, with coefficients denoted by $A_a$: $$T^{ij} = A_a \mathcal{J}^{ij}_a.$$

This object is an N×N antisymmetric matrix, but alsoalso an $\frac{1}{2}N(N-1)$-dimensional vector in the adjoint representation vector space, introduced above, itching to be rotated by the suitable    $\frac{1}{2}N(N-1)\times\frac{1}{2}N(N-1)$ adjoint representation matrices.

The dimension of the adjoint representation is also (These matrices are gotten/furnished by the dimensionstructure constants of the corresponding Lie algebra you'd get from your (regarded as a vector space as you saw)$\cal J$s in the D=N irrep., but don't worry about them.)

  • The dimension of the adjoint representation is also the dimension of the corresponding Lie algebra (regarded as a vector space as you saw).

I don't have that text, but your question is standard stuff. I suspect what you are missing is the fast interplay of D-dimensional vectors and the D×D matrices acting on them, sometimes regarded as vector elements themselves, of a larger vector!

The dimension D of the representation is the dimensionality of the "vectors" the D×D representation matrices act on. So, all kinds of objects "furnish" D-dimensional vectors on which representation matrices act to transform.

Antisymmetric tensors $T^{ij}$ in $SO(N)$ with indices $j=1,...,N$ comprise a set of $\frac{1}{2}N(N-1)$ independent matrices, so they furnish vectors of a D-dimensional representation, with $D=\frac{1}{2}N(N-1)$. Each such matrix is a component of the D-vector.

It is also true that there are $\frac{1}{2}N(N-1)$ generators of $SO(N)$, indexed by the (mn) pairs, $$\mathcal{J}_{(m n)}^{i j}=\left(\delta^{m i} \delta^{n j}-\delta^{m j} \delta^{n i}\right)$$ which act on the space of N-dimensional js to rotate them to such is.

  • However, these matrices may also be regarded as the $D=\frac{1}{2}N(N-1)$ components of a new vector, acted upon by some $\frac{1}{2}N(N-1) \times \frac{1}{2}N(N-1)$ matrices, ( actually $\frac{1}{2}N(N-1)$ independent such, indexed by (mn). This is the adjoint representation.)

So the D=N generators became the $D=\frac{1}{2}N(N-1)$-vector elements of the adjoint. Fortuitously, these two dimensions are the same for N=3, so the fundamental of SO(3) is also the adjoint!

Now

We can also regard the antisymmetric tensor $T^{ij}$ as an $N$-by-$N$ matrix, and hence write it as a linear combination of the $\mathcal{J}_a$'s, with coefficients denoted by $A_a$: $$T^{ij} = A_a \mathcal{J}^{ij}_a.$$

This object is an N×N antisymmetric matrix, but also an $\frac{1}{2}N(N-1)$-dimensional vector in the adjoint representation vector space, introduced above, itching to be rotated by the suitable  $\frac{1}{2}N(N-1)\times\frac{1}{2}N(N-1)$ adjoint representation matrices.

The dimension of the adjoint representation is also the dimension of the corresponding Lie algebra (regarded as a vector space as you saw).

I don't have that text, but your question is standard stuff. I suspect what you are missing is the fast interplay of D-dimensional vectors and the D×D matrices acting on them, sometimes regarded as vector elements themselves, of a larger vector!

The dimension D of the representation is the dimensionality of the "vectors" on which the D×D representation matrices act. But all kinds of objects "furnish" D-dimensional vectors on which the suitable representation matrices act to transform them. "Furnish" means provide components of a large D-dimensional vector specifying a possibly new representation.

Antisymmetric tensors $T^{ij}$ in $SO(N)$ with indices $j=1,...,N$ comprise a set of $\frac{1}{2}N(N-1)$ independent antisymmetric matrices, so they furnish vectors of an irreducible D-dimensional representation, with $D=\frac{1}{2}N(N-1)$: they define the relevant vector space. Each such matrix is a component of the D-vector.

It is also true that there are $\frac{1}{2}N(N-1)$ generators of $SO(N)$, indexed by the (mn) pairs, $$\mathcal{J}_{(m n)}^{i j}=\left(\delta^{m i} \delta^{n j}-\delta^{m j} \delta^{n i}\right)$$ which act on the space of N-dimensional js to rotate them to such is.

  • However, these N×N antisymmetric matrices may also be regarded as the $D=\frac{1}{2}N(N-1)$ components of a new vector, acted upon by some larger $\frac{1}{2}N(N-1) \times \frac{1}{2}N(N-1)$ matrices, (actually $\frac{1}{2}N(N-1)$ independent such, indexed by (mn). This is the adjoint representation.)

So the D=N generators became the $D=\frac{1}{2}N(N-1)$-vector elements of the adjoint (which they furnished). Fortuitously, these two dimensions are the same for N=3, so the fundamental of SO(3) happens to also be its adjoint!

Now

We can also regard the antisymmetric tensor $T^{ij}$ as an $N$-by-$N$ matrix, and hence write it as a linear combination of the $\mathcal{J}_a$'s, with coefficients denoted by $A_a$: $$T^{ij} = A_a \mathcal{J}^{ij}_a.$$

This object is an N×N antisymmetric matrix, but also an $\frac{1}{2}N(N-1)$-dimensional vector in the adjoint representation vector space, introduced above, itching to be rotated by the suitable  $\frac{1}{2}N(N-1)\times\frac{1}{2}N(N-1)$ adjoint representation matrices. (These matrices are gotten/furnished by the structure constants of the Lie algebra you'd get from your $\cal J$s in the D=N irrep., but don't worry about them.)

  • The dimension of the adjoint representation is also the dimension of the corresponding Lie algebra (regarded as a vector space as you saw).
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Cosmas Zachos
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I don't have that text, but your question is standard stuff. I suspect what you are missing is the fast interplay of D-dimensional vectors and the D×D matrices acting on them, sometimes regarded as vector elements themselves, of a larger vector!

The dimension D of the representation is the dimensionality of the "vectors" the D×D representation matrices act on. So, all kinds of objects "furnish" D-dimensional vectors on which representation matrices act to transform.

Antisymmetric tensors $T^{ij}$ in $SO(N)$ with indices $j=1,...,N$ comprise a set of $\frac{1}{2}N(N-1)$ independent matrices, so they furnish vectors of a D-dimensional representation, with $D=\frac{1}{2}N(N-1)$. Each such matrix is a component of the D-vector.

It is also true that there are $\frac{1}{2}N(N-1)$ generators of $SO(N)$, indexed by the (mn) pairs, $$\mathcal{J}_{(m n)}^{i j}=\left(\delta^{m i} \delta^{n j}-\delta^{m j} \delta^{n i}\right)$$ which act on the space of N-dimensional js to rotate them to such is.

  • However, these matrices may also be regarded as the $D=\frac{1}{2}N(N-1)$ components of a new vector, acted upon by some $\frac{1}{2}N(N-1) \times \frac{1}{2}N(N-1)$ matrices, ( actually $\frac{1}{2}N(N-1)$ independent such, indexed by (mn). This is the adjoint representation.)

However, these matrices may also be regarded as the $D=\frac{1}{2}N(N-1)$ components of a new vector, acted upon by some $\frac{1}{2}N(N-1) \times \frac{1}{2}N(N-1)$ matrices, $\frac{1}{2}N(N-1)$ independent such, indexed by (mn). SoSo the D=N generators becomebecame the $D=\frac{1}{2}N(N-1)$-vector elements of the adjoint. Fortuitously, these two dimensions are the same for N=3, so the fundamental of SO(3) isis also the adjoint!

Now

We can also regard the antisymmetric tensor $T^{ij}$ as an $N$-by-$N$ matrix, and hence write it as a linear combination of the $\mathcal{J}_a$'s, with coefficients denoted by $A_a$: $$T^{ij} = A_a \mathcal{J}^{ij}_a.$$

This object is an N×N antisymmetric matrix, but also an $\frac{1}{2}N(N-1)$-dimensional vector in the adjoint representation vector space, introduced above, itching to be rotated by the suitable $\frac{1}{2}N(N-1)\times\frac{1}{2}N(N-1)$ adjoint representation matrices.

The dimension of the adjoint representation is also the dimension of the corresponding Lie algebra (regarded as a vector space as you saw).

I don't have that text, but your question is standard stuff. I suspect what you are missing is the fast interplay of D-dimensional vectors and the D×D matrices acting on them, sometimes regarded as vector elements themselves, of a larger vector!

The dimension D of the representation is the dimensionality of the "vectors" the D×D representation matrices act on. So, all kinds of objects "furnish" D-dimensional vectors on which representation matrices act to transform.

Antisymmetric tensors $T^{ij}$ in $SO(N)$ with indices $j=1,...,N$ comprise a set of $\frac{1}{2}N(N-1)$ independent matrices, so they furnish vectors of a D-dimensional representation, with $D=\frac{1}{2}N(N-1)$. Each such matrix is a component of the D-vector.

It is also true that there are $\frac{1}{2}N(N-1)$ generators of $SO(N)$, indexed by the (mn) pairs, $$\mathcal{J}_{(m n)}^{i j}=\left(\delta^{m i} \delta^{n j}-\delta^{m j} \delta^{n i}\right)$$ which act on the space of N-dimensional js to rotate them to such is.

However, these matrices may also be regarded as the $D=\frac{1}{2}N(N-1)$ components of a new vector, acted upon by some $\frac{1}{2}N(N-1) \times \frac{1}{2}N(N-1)$ matrices, $\frac{1}{2}N(N-1)$ independent such, indexed by (mn). So the D=N generators become the $D=\frac{1}{2}N(N-1)$-vector elements. Fortuitously, these two dimensions are the same for N=3, so the fundamental of SO(3) is also the adjoint!

Now

We can also regard the antisymmetric tensor $T^{ij}$ as an $N$-by-$N$ matrix, and hence write it as a linear combination of the $\mathcal{J}_a$'s, with coefficients denoted by $A_a$: $$T^{ij} = A_a \mathcal{J}^{ij}_a.$$

This object is an N×N antisymmetric matrix, but also an $\frac{1}{2}N(N-1)$-dimensional vector in the adjoint representation vector space, itching to be rotated by the suitable $\frac{1}{2}N(N-1)\times\frac{1}{2}N(N-1)$ adjoint representation matrices.

I don't have that text, but your question is standard stuff. I suspect what you are missing is the fast interplay of D-dimensional vectors and the D×D matrices acting on them, sometimes regarded as vector elements themselves, of a larger vector!

The dimension D of the representation is the dimensionality of the "vectors" the D×D representation matrices act on. So, all kinds of objects "furnish" D-dimensional vectors on which representation matrices act to transform.

Antisymmetric tensors $T^{ij}$ in $SO(N)$ with indices $j=1,...,N$ comprise a set of $\frac{1}{2}N(N-1)$ independent matrices, so they furnish vectors of a D-dimensional representation, with $D=\frac{1}{2}N(N-1)$. Each such matrix is a component of the D-vector.

It is also true that there are $\frac{1}{2}N(N-1)$ generators of $SO(N)$, indexed by the (mn) pairs, $$\mathcal{J}_{(m n)}^{i j}=\left(\delta^{m i} \delta^{n j}-\delta^{m j} \delta^{n i}\right)$$ which act on the space of N-dimensional js to rotate them to such is.

  • However, these matrices may also be regarded as the $D=\frac{1}{2}N(N-1)$ components of a new vector, acted upon by some $\frac{1}{2}N(N-1) \times \frac{1}{2}N(N-1)$ matrices, ( actually $\frac{1}{2}N(N-1)$ independent such, indexed by (mn). This is the adjoint representation.)

So the D=N generators became the $D=\frac{1}{2}N(N-1)$-vector elements of the adjoint. Fortuitously, these two dimensions are the same for N=3, so the fundamental of SO(3) is also the adjoint!

Now

We can also regard the antisymmetric tensor $T^{ij}$ as an $N$-by-$N$ matrix, and hence write it as a linear combination of the $\mathcal{J}_a$'s, with coefficients denoted by $A_a$: $$T^{ij} = A_a \mathcal{J}^{ij}_a.$$

This object is an N×N antisymmetric matrix, but also an $\frac{1}{2}N(N-1)$-dimensional vector in the adjoint representation vector space, introduced above, itching to be rotated by the suitable $\frac{1}{2}N(N-1)\times\frac{1}{2}N(N-1)$ adjoint representation matrices.

The dimension of the adjoint representation is also the dimension of the corresponding Lie algebra (regarded as a vector space as you saw).

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Cosmas Zachos
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I don't have that text, but your question is standard stuff. I suspect what you are missing is the fast interplay of D-dimensional vectors and the DxDD×D matrices acting on them, sometimes regarded as vector elements themselves, of a larger vector!

The dimension D of the representation is the dimensionality of the "vectors" the D×D representation matrices act on. So, all kinds of objects "furnish" D-dimensional vectors on which representation matrices act to transform.

Antisymmetric tensors $T^{ij}$ in $SO(N)$ with indices $j=1,...,N$ comprise a set of $\frac{1}{2}N(N-1)$ independent matrices, so they furnish vectors of a D-dimensional representation, with $D=\frac{1}{2}N(N-1)$. Each such matrix is a component of the D-vector.

It is also true that there are $\frac{1}{2}N(N-1)$ generators of $SO(N)$, indexed by the (mn) pairs, $$\mathcal{J}_{(m n)}^{i j}=\left(\delta^{m i} \delta^{n j}-\delta^{m j} \delta^{n i}\right)$$ which act on the space of N-dimensional js to rotate them to such is.

However, these matrices may also be regarded as the $D=\frac{1}{2}N(N-1)$ components of a new vector, acted upon by some $\frac{1}{2}N(N-1) \times \frac{1}{2}N(N-1)$ matrices, $\frac{1}{2}N(N-1)$ independent such, indexed by (mn). So the D=N generators become the $D=\frac{1}{2}N(N-1)$-vector elements. Fortuitously, these two dimensions are the same for N=3, so the fundamental of SO(3) is also the adjoint!

Now

We can also regard the antisymmetric tensor $T^{ij}$ as an $N$-by-$N$ matrix, and hence write it as a linear combination of the $\mathcal{J}_a$'s, with coefficients denoted by $A_a$: $$T^{ij} = A_a \mathcal{J}^{ij}_a.$$

This object is an N×N antisymmetric matrix, but also an $\frac{1}{2}N(N-1)$-dimensional vector in the adjoint representation vector space, itching to be rotated by the suitable $\frac{1}{2}N(N-1)\times\frac{1}{2}N(N-1)$ adjoint representation matrices.

I don't have that text, but your question is standard stuff. I suspect what you are missing is the fast interplay of D-dimensional vectors and the DxD matrices acting on them, sometimes regarded as vector elements themselves, of a larger vector!

The dimension D of the representation is the dimensionality of the "vectors" the D×D representation matrices act on. So, all kinds of objects "furnish" D-dimensional vectors on which representation matrices act to transform.

Antisymmetric tensors $T^{ij}$ in $SO(N)$ with indices $j=1,...,N$ comprise a set of $\frac{1}{2}N(N-1)$ independent matrices, so they furnish vectors of a D-dimensional representation, with $D=\frac{1}{2}N(N-1)$. Each such matrix is a component of the D-vector.

It is also true that there are $\frac{1}{2}N(N-1)$ generators of $SO(N)$, indexed by the (mn) pairs, $$\mathcal{J}_{(m n)}^{i j}=\left(\delta^{m i} \delta^{n j}-\delta^{m j} \delta^{n i}\right)$$ which act on the space of N-dimensional js to rotate them to such is.

However, these matrices may also be regarded as the $D=\frac{1}{2}N(N-1)$ components of a new vector, acted upon by some $\frac{1}{2}N(N-1) \times \frac{1}{2}N(N-1)$ matrices, $\frac{1}{2}N(N-1)$ independent such, indexed by (mn). So the D=N generators become the $D=\frac{1}{2}N(N-1)$-vector elements. Fortuitously, these two dimensions are the same for N=3, so the fundamental of SO(3) is also the adjoint!

Now

We can also regard the antisymmetric tensor $T^{ij}$ as an $N$-by-$N$ matrix, and hence write it as a linear combination of the $\mathcal{J}_a$'s, with coefficients denoted by $A_a$: $$T^{ij} = A_a \mathcal{J}^{ij}_a.$$

This object is an N×N antisymmetric matrix, but also an $\frac{1}{2}N(N-1)$-dimensional vector in the adjoint representation vector space, itching to be rotated by the suitable $\frac{1}{2}N(N-1)\times\frac{1}{2}N(N-1)$ adjoint representation matrices.

I don't have that text, but your question is standard stuff. I suspect what you are missing is the fast interplay of D-dimensional vectors and the D×D matrices acting on them, sometimes regarded as vector elements themselves, of a larger vector!

The dimension D of the representation is the dimensionality of the "vectors" the D×D representation matrices act on. So, all kinds of objects "furnish" D-dimensional vectors on which representation matrices act to transform.

Antisymmetric tensors $T^{ij}$ in $SO(N)$ with indices $j=1,...,N$ comprise a set of $\frac{1}{2}N(N-1)$ independent matrices, so they furnish vectors of a D-dimensional representation, with $D=\frac{1}{2}N(N-1)$. Each such matrix is a component of the D-vector.

It is also true that there are $\frac{1}{2}N(N-1)$ generators of $SO(N)$, indexed by the (mn) pairs, $$\mathcal{J}_{(m n)}^{i j}=\left(\delta^{m i} \delta^{n j}-\delta^{m j} \delta^{n i}\right)$$ which act on the space of N-dimensional js to rotate them to such is.

However, these matrices may also be regarded as the $D=\frac{1}{2}N(N-1)$ components of a new vector, acted upon by some $\frac{1}{2}N(N-1) \times \frac{1}{2}N(N-1)$ matrices, $\frac{1}{2}N(N-1)$ independent such, indexed by (mn). So the D=N generators become the $D=\frac{1}{2}N(N-1)$-vector elements. Fortuitously, these two dimensions are the same for N=3, so the fundamental of SO(3) is also the adjoint!

Now

We can also regard the antisymmetric tensor $T^{ij}$ as an $N$-by-$N$ matrix, and hence write it as a linear combination of the $\mathcal{J}_a$'s, with coefficients denoted by $A_a$: $$T^{ij} = A_a \mathcal{J}^{ij}_a.$$

This object is an N×N antisymmetric matrix, but also an $\frac{1}{2}N(N-1)$-dimensional vector in the adjoint representation vector space, itching to be rotated by the suitable $\frac{1}{2}N(N-1)\times\frac{1}{2}N(N-1)$ adjoint representation matrices.

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Cosmas Zachos
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