We have just seen the decomposition
$$ \mathfrak{o}(V\oplus V^*) = \textrm{End}(V)\oplus \wedge^2 V^* \oplus \wedge^2 V.$$
By looking at these elements as matrices, we have
$$ \left( \begin{array}{cc} A & \beta \\ B & -A^T \end{array} \right),$$
with \( A \in \textrm{End}(V) \), \( B\in \wedge^2 V^* \), \( \beta \in \wedge^2 V \).
Do not forget that we did this because we wanted to talk about \( \textrm{O}(V\oplus V^*) \). The way we have to pass information from the Lie algebra to the Lie group is the exponential. Since we are dealing with groups of matrices, the exponential is just the usual exponential \( e^A = \sum_{i=0}^{+\infty} \frac{A^i}{i!} \).
If we take \( A\in \textrm{End}(V) \), we get the element
$$ \left( \begin{array}{cc} e^A & 0 \\ 0 & ((e^A)^{*})^{-1} \end{array} \right) = \left( \begin{array}{cc} P & 0 \\ 0 & ((P)^{*})^{-1} \end{array} \right) \in \textrm{O}(V\oplus V^*), $$
where \(P \in \textrm{GL}^+(V) \) ( \(\textrm{det}P>0 \) ), although if the determinant of \( P \) is negative, the matrix you can see is clearly still in \( \textrm{O}(V\oplus V^*) \). This is showing that, as we could expect, the generalized symmetries contain the classical symmetries.
Let us look for something new. Starting with \( B\in \wedge^2 V^* \) or \( \beta \in \wedge^2 V \) we get the \( B \)-fields and the \( \beta \)-fields:
$$ \left( \begin{array}{cc} 1 & 0 \\ B & 1 \end{array} \right), \qquad \qquad \qquad \left( \begin{array}{cc} 1 & \beta \\ 0 & 1 \end{array} \right).$$
Notice that the \( B \)-fields, which act on \( X+\xi \in V\oplus V^* \) by $$X+\xi \mapsto X+\xi+i_X B,$$ preserve the projection to \( V \). We will soon forget about \( \beta \) -fields and focus on \(B \)-fields.
$$ \mathfrak{o}(V\oplus V^*) = \textrm{End}(V)\oplus \wedge^2 V^* \oplus \wedge^2 V.$$
By looking at these elements as matrices, we have
$$ \left( \begin{array}{cc} A & \beta \\ B & -A^T \end{array} \right),$$
with \( A \in \textrm{End}(V) \), \( B\in \wedge^2 V^* \), \( \beta \in \wedge^2 V \).
Do not forget that we did this because we wanted to talk about \( \textrm{O}(V\oplus V^*) \). The way we have to pass information from the Lie algebra to the Lie group is the exponential. Since we are dealing with groups of matrices, the exponential is just the usual exponential \( e^A = \sum_{i=0}^{+\infty} \frac{A^i}{i!} \).
If we take \( A\in \textrm{End}(V) \), we get the element
$$ \left( \begin{array}{cc} e^A & 0 \\ 0 & ((e^A)^{*})^{-1} \end{array} \right) = \left( \begin{array}{cc} P & 0 \\ 0 & ((P)^{*})^{-1} \end{array} \right) \in \textrm{O}(V\oplus V^*), $$
where \(P \in \textrm{GL}^+(V) \) ( \(\textrm{det}P>0 \) ), although if the determinant of \( P \) is negative, the matrix you can see is clearly still in \( \textrm{O}(V\oplus V^*) \). This is showing that, as we could expect, the generalized symmetries contain the classical symmetries.
Let us look for something new. Starting with \( B\in \wedge^2 V^* \) or \( \beta \in \wedge^2 V \) we get the \( B \)-fields and the \( \beta \)-fields:
$$ \left( \begin{array}{cc} 1 & 0 \\ B & 1 \end{array} \right), \qquad \qquad \qquad \left( \begin{array}{cc} 1 & \beta \\ 0 & 1 \end{array} \right).$$
Notice that the \( B \)-fields, which act on \( X+\xi \in V\oplus V^* \) by $$X+\xi \mapsto X+\xi+i_X B,$$ preserve the projection to \( V \). We will soon forget about \( \beta \) -fields and focus on \(B \)-fields.
No comments:
Post a Comment
Please, use the comments to share any thoughts or concerns, and also to leave some anonymous feedback, which will not be published.