In the case of the root system $A_{n-1}$, I want to understand the correspondence between the Lie group and the Lie algebra. Please help me understand the relationship between the Lie groups $SU(n)$ and $SL(n)$ and the Lie algebra $\mathfrak{sl}(n,\mathbb{C})$.

This is what I think I understand. The special unitary group $SU(n)$ is the Lie group of $n\times n$ unitary matrices with determinant $1$. A matrix is unitary if its conjugate transpose is also its inverse: $UU^*=U^*U=I$. The rows form an orthonormal basis of $\mathbb{C}^n$, and so do the columns, and the rows and columns are orthonormal with respect to each other.

There is a correspondence between a Lie group $G$ and its Lie algebra $\mathrm{Lie}(G)$ given by:

$\mathrm{Lie}(G)= \{ X\in M_n(\mathbb{C})|e^{tX}\in G \textrm{ for all }t\in \mathbb{R} \}$

The formula $\mathrm {det}(e^X)=e^{\mathrm {tr} (X)}$ and the condition that the $\mathrm{det}(U)=1$ for all $U\in SU(n)$ means that $\mathrm{tr}(X)=0$ for all $X\in \mathrm{Lie}(SU(n))$. Which is to say, the Lie algebra $\mathrm{Lie}(SU(n))$ is a subset of the special linear algebra $\mathfrak{sl}(n)=\{X\in M_n(\mathbb{C})|\mathrm{tr}(X)=0\}$.

The condition $UU^*=I$ for $U\in SU(n)$ translates into the condition $X-X^*=0$ for $X\in \mathrm{Lie}(SU(n))$. This means that $x_{ij}=\overline{x}_{ji}$, and in particular, $x_{ii}$ is purely imaginary. These are the skew Hermitian matrices.

Combining the two conditions, I think we should have that $\mathrm{Lie}(SU(n))$ is the set of skew Hermitian $n \times n$ matrices in $M_n(\mathbb{C})$ with zero trace. That would be perhaps $\mathfrak{su}(n,\mathbb{C})$.

But the Lie algebra associated with the root system $A_{n-1}$ is given by $\mathfrak{sl}(n,\mathbb{C})$. I suppose that makes sense given from Victor Kac's Lecture 14 that the related Cartan subalgebra $\mathfrak{h}$ is the set of all traceless diagonal matrices.

I see at the Wikipedia article on Lie group-Lie algebra correspondence that for root system $A_n$, the compact Lie group is $SU(n+1)$ and the complexification of the associated Lie algebra is $\mathfrak{sl}(n+1)$. I don't understand the relevance here of complexification but I imagine that may be what I'm not understanding.

Please, what is the connection between $SU(n)$, $SL(n)$, $\mathfrak{sl}(n)$ and $A_{n-1}$?

I would also appreciate an explanation of what this means in the cases $n=2$ and $n=3$.

Thank you!

In the case of the root system $A_{n-1}$, I want to understand the correspondence between the Lie group and the Lie algebra. Please help me understand the relationship between the Lie groups $SU(n)$ and $SL(n)$ and the Lie algebra $\mathfrak{sl}(n,\mathbb{C})$.

This is what I think I understand. The special unitary group $SU(n)$ is the Lie group of $n\times n$ unitary matrices with determinant $1$. A matrix is unitary if its conjugate transpose is also its inverse: $UU^*=U^*U=I$. The rows form an orthonormal basis of $\mathbb{C}^n$, and so do the columns, and the rows and columns are orthonormal with respect to each other.

There is a correspondence between a Lie group $G$ and its Lie algebra $\mathrm{Lie}(G)$ given by:

$\mathrm{Lie}(G)= \{ X\in M_n(\mathbb{C})|e^{tX}\in G \textrm{ for all }t\in \mathbb{R} \}$

The formula $\mathrm {det}(e^X)=e^{\mathrm {tr} (X)}$ and the condition that the $\mathrm{det}(U)=1$ for all $U\in SU(n)$ means that $\mathrm{tr}(X)=0$ for all $X\in \mathrm{Lie}(SU(n))$. Which is to say, the Lie algebra $\mathrm{Lie}(SU(n))$ is a subset of the special linear algebra $\mathfrak{sl}(n)=\{X\in M_n(\mathbb{C})|\mathrm{tr}(X)=0\}$.

The condition $UU^*=I$ for $U\in SU(n)$ translates into the condition $X-X^*=0$ for $X\in \mathrm{Lie}(SU(n))$. This means that $x_{ij}=\overline{x}_{ji}$, and in particular, $x_{ii}$ is purely imaginary. These are the skew Hermitian matrices.

Combining the two conditions, I think we should have that $\mathrm{Lie}(SU(n))$ is the set of skew Hermitian $n \times n$ matrices in $M_n(\mathbb{C})$ with zero trace. That would be perhaps $\mathfrak{su}(n,\mathbb{C})$.

But the Lie algebra associated with the root system $A_{n-1}$ is given by $\mathfrak{sl}(n,\mathbb{C})$. I suppose that makes sense given from Victor Kac's Lecture 14 that the related Cartan subalgebra $\mathfrak{h}$ is the set of all traceless diagonal matrices.

I see at the Wikipedia article on Lie group-Lie algebra correspondence that for root system $A_n$, the compact Lie group is $SU(n+1)$ and the complexification of the associated Lie algebra is $\mathfrak{sl}(n+1)$. I don't understand the relevance here of complexification but I imagine that may be what I'm not understanding.

The table at the Wikipedia article on Classical group lists, for the root system $A_{n-1}$, with a complex Lie algebra, the group $SL(n,\mathbb{C})$ and the maximal compact subgroup as $SU(n)$. So I suppose an issue here is that the corresponding group would be $SL(n,\mathbb{C})$, but it is not compact, so we need to restrict ourselves to the maximal compact subgroup. Even if that's the case, I still need to understand why, starting from $SU(n)$, the correspondence doesn't take us to $\mathfrak{su}$. Is that not semi-simple? Is that the problem?

Please, what is the connection between $SU(n)$, $SL(n)$, $\mathfrak{sl}(n)$ and $A_{n-1}$?

I would also appreciate an explanation of what this means in the cases $n=2$ and $n=3$.

Thank you!

In the case of the root system $A_{n-1}$, I want to understand the correspondence between the Lie group and the Lie algebra. Please help me understand the relationship between the Lie groups $SU(n)$ and $SL(n)$ and the Lie algebra $\frak{sl}$$(n,\mathbb{C})$.

This is what I think I understand. The special unitary group $SU(n)$ is the Lie group of $n\times n$ unitary matrices with determinant $1$. A matrix is unitary if its conjugate transpose is also its inverse: $UU^*=U^*U=I$. The rows form an orthonormal basis of $\mathbb{C}^n$, and so do the columns, and the rows and columns are orthonormal with respect to each other.

There is a correspondence between a Lie group $G$ and its Lie algebra $\mathrm{Lie}(G)$ given by:

$\mathrm{Lie}(G)= \{ X\in M_n(\mathbb{C})|e^{tX}\in G \textrm{ for all }t\in \mathbb{R} \}$

The formula $\mathrm {det}(e^X)=e^{\mathrm {tr} (X)}$ and the condition that the $\mathrm{det}(U)=1$ for all $U\in SU(n)$ means that $\mathrm{tr}(X)=0$ for all $X\in \mathrm{Lie}(SU(n))$. Which is to say, the Lie algebra $\mathrm{Lie}(SU(n))$ is a subset of the special linear algebra $\mathfrak{sl}(n)=\{X\in M_n(\mathbb{C})|\mathrm{tr}(X)=0\}$.

The condition $UU^*=I$ for $U\in SU(n)$ translates into the condition $X-X^*=0$ for $X\in \mathrm{Lie}(SU(n))$. This means that $x_{ij}=\overline{x}_{ji}$, and in particular, $x_{ii}$ is purely imaginary. These are the skew Hermitian matrices.

Combining the two conditions, I think we should have that $\mathrm{Lie}(SU(n))$ is the set of skew Hermitian $n \times n$ matrices in $M_n(\mathbb{C})$ with zero trace. That would be perhaps $\frak{su}$$(n,\mathbb{C})$.

But the Lie algebra associated with the root system $A_{n-1}$ is given by $\frak{sl}$$(n,\mathbb{C})$. I suppose that makes sense given from Victor Kac's Lecture 14 that the related Cartan subalgebra $\frak{h}$ is the set of all traceless diagonal matrices.

I see at the Wikipedia article on Lie group-Lie algebra correspondence that for root system $A_n$, the compact Lie group is $SU(n+1)$ and the complexification of the associated Lie algebra is $\frak{sl}(n+1)$. I don't understand the relevance here of complexification but I imagine that may be what I'm not understanding.

Please, what is the connection between $SU(n)$, $SL(n)$, $\frak{sl}(n)$ and $A_{n-1}$?

I would also appreciate an explanation of what this means in the cases $n=2$ and $n=3$.

Thank you!

In the case of the root system $A_{n-1}$, I want to understand the correspondence between the Lie group and the Lie algebra. Please help me understand the relationship between the Lie groups $SU(n)$ and $SL(n)$ and the Lie algebra $\mathfrak{sl}(n,\mathbb{C})$.

This is what I think I understand. The special unitary group $SU(n)$ is the Lie group of $n\times n$ unitary matrices with determinant $1$. A matrix is unitary if its conjugate transpose is also its inverse: $UU^*=U^*U=I$. The rows form an orthonormal basis of $\mathbb{C}^n$, and so do the columns, and the rows and columns are orthonormal with respect to each other.

There is a correspondence between a Lie group $G$ and its Lie algebra $\mathrm{Lie}(G)$ given by:

$\mathrm{Lie}(G)= \{ X\in M_n(\mathbb{C})|e^{tX}\in G \textrm{ for all }t\in \mathbb{R} \}$

The formula $\mathrm {det}(e^X)=e^{\mathrm {tr} (X)}$ and the condition that the $\mathrm{det}(U)=1$ for all $U\in SU(n)$ means that $\mathrm{tr}(X)=0$ for all $X\in \mathrm{Lie}(SU(n))$. Which is to say, the Lie algebra $\mathrm{Lie}(SU(n))$ is a subset of the special linear algebra $\mathfrak{sl}(n)=\{X\in M_n(\mathbb{C})|\mathrm{tr}(X)=0\}$.

The condition $UU^*=I$ for $U\in SU(n)$ translates into the condition $X-X^*=0$ for $X\in \mathrm{Lie}(SU(n))$. This means that $x_{ij}=\overline{x}_{ji}$, and in particular, $x_{ii}$ is purely imaginary. These are the skew Hermitian matrices.

Combining the two conditions, I think we should have that $\mathrm{Lie}(SU(n))$ is the set of skew Hermitian $n \times n$ matrices in $M_n(\mathbb{C})$ with zero trace. That would be perhaps $\mathfrak{su}(n,\mathbb{C})$.

But the Lie algebra associated with the root system $A_{n-1}$ is given by $\mathfrak{sl}(n,\mathbb{C})$. I suppose that makes sense given from Victor Kac's Lecture 14 that the related Cartan subalgebra $\mathfrak{h}$ is the set of all traceless diagonal matrices.

I see at the Wikipedia article on Lie group-Lie algebra correspondence that for root system $A_n$, the compact Lie group is $SU(n+1)$ and the complexification of the associated Lie algebra is $\mathfrak{sl}(n+1)$. I don't understand the relevance here of complexification but I imagine that may be what I'm not understanding.

Please, what is the connection between $SU(n)$, $SL(n)$, $\mathfrak{sl}(n)$ and $A_{n-1}$?

I would also appreciate an explanation of what this means in the cases $n=2$ and $n=3$.

Thank you!