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Asaf
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It is evident that the singular vectors are defined as the ``$u_{A}$-part which is $g_{t}$ divergent in the future'', this gives $m\cdot n$ ($=\dim \left(u_{A}\right)$) minus the dimension of the singular vectors.

The question here is what's going on with the ''extra dimensions''. Notice that $u_{A}$ is a unstable horospherical subgroup of $g_{t}$.

By definition you may consider its Lie algebra as the part of the $Lie(G)$ for which $Ad(g_{1})$ is expanding (notice that $Ad(g)$ is indeed diagonalizable).

For the ''other directions'', as $Ad(g_{t})$ is not expanding, assume you as given something like $\exp(\underline{h}).\mathbb{Z}^{m+n}$ then you may write $g_{t}.\exp(\underline{h}).\mathbb{Z}^{m+n} = \exp(Ad(g_{t}).\underline{h})g_{t}.\mathbb{Z}^{m+n}$. Notice that $g_{t}.\mathbb{Z}^{m+n}$ is diverging, and $$ dist(\exp(Ad(g_{t}).\underline{h})g_{t}.\mathbb{Z}^{m+n},g_{t}.\mathbb{Z}^{m+n}) \ll_{h} 1$$ due to the fact that $Ad(g_{t})$ is not expanding $\underline{h}$.

It is evident that the singular vectors are defined as the ``$u_{A}$-part which is $g_{t}$ divergent in the future'', this gives $m\cdot n$ ($=\dim \left(u_{A}\right)$) minus the dimension of the singular vectors.

The question here is what's going on with the ''extra dimensions''. Notice that $u_{A}$ is a unstable horospherical subgroup of $g_{t}$.

By definition you may consider its Lie algebra as the part of the $Lie(G)$ for which $Ad(g_{1})$ is expanding (notice that $Ad(g)$ is indeed diagonalizable).

For the ''other directions'', as $Ad(g_{t})$ is not expanding, assume you as given something like $\exp(\underline{h}).\mathbb{Z}^{m+n}$ then you may write $g_{t}.\exp(\underline{h}).\mathbb{Z}^{m+n} = \exp(Ad(g_{t}).\underline{h})g_{t}.\mathbb{Z}^{m+n}$. Notice that $g_{t}.\mathbb{Z}^{m+n}$ is diverging, and $$ dist(\exp(Ad(g_{t}).\underline{h})g_{t}.\mathbb{Z}^{m+n},g_{t}.\mathbb{Z}^{m+n}) \ll_{h} 1$$ due to the fact that $Ad(g_{t})$ is not expanding.

It is evident that the singular vectors are defined as the ``$u_{A}$-part which is $g_{t}$ divergent in the future'', this gives $m\cdot n$ ($=\dim \left(u_{A}\right)$) minus the dimension of the singular vectors.

The question here is what's going on with the ''extra dimensions''. Notice that $u_{A}$ is a unstable horospherical subgroup of $g_{t}$.

By definition you may consider its Lie algebra as the part of the $Lie(G)$ for which $Ad(g_{1})$ is expanding (notice that $Ad(g)$ is indeed diagonalizable).

For the ''other directions'', as $Ad(g_{t})$ is not expanding, assume you as given something like $\exp(\underline{h}).\mathbb{Z}^{m+n}$ then you may write $g_{t}.\exp(\underline{h}).\mathbb{Z}^{m+n} = \exp(Ad(g_{t}).\underline{h})g_{t}.\mathbb{Z}^{m+n}$. Notice that $g_{t}.\mathbb{Z}^{m+n}$ is diverging, and $$ dist(\exp(Ad(g_{t}).\underline{h})g_{t}.\mathbb{Z}^{m+n},g_{t}.\mathbb{Z}^{m+n}) \ll_{h} 1$$ due to the fact that $Ad(g_{t})$ is not expanding $\underline{h}$.

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Asaf
  • 2.5k
  • 20
  • 19

It is evident that the singular vectors are defined as the ``$u_{A}$-part which is $g_{t}$ divergent in the future'', this gives $m\cdot n$ ($=\dim \left(u_{A}\right)$) minus the dimension of the singular vectors.

The question here is what's going on with the ''extra dimensions''. Notice that $u_{A}$ is a unstable horospherical subgroup of $g_{t}$.

By definition you may consider its Lie algebra as the part of the $Lie(G)$ for which $Ad(g_{1})$ is expanding (notice that $Ad(g)$ is indeed diagonalizable).

For the ''other directions'', as $Ad(g_{t})$ is not expanding, assume you as given something like $\exp(\underline{h}).\mathbb{Z}^{m+n}$ then you may write $g_{t}.\exp(\underline{h}).\mathbb{Z}^{m+n} = \exp(Ad(g_{t}).\underline{h})g_{t}.\mathbb{Z}^{m+n}$. Notice that $g_{t}.\mathbb{Z}^{m+n}$ is diverging, and $$ dist(\exp(Ad(g_{t}).\underline{h})g_{t}.\mathbb{Z}^{m+n},g_{t}.\mathbb{Z}^{m+n}) \ll_{h} 1$$ due to the fact that $Ad(g_{t})$ is not expanding.