# About calculating a fundamental group

• A
• aalma
In summary, in order to compute ##\pi_1(PGL_2(R))##, one can use the fact that there is a bundle ##\mathbb{R}^\times\to GL_n(\mathbb{R})\to PGL_n(\mathbb{R})## and the identity component of the fiber is contractible. This means that ##PGL_n(\mathbb{R})## and ##GL_n(\mathbb{R})## have the same homotopy groups in positive degrees, and the identity component of ##GL_n(\mathbb{R})## is homotopy equivalent to ##SO(n;\mathbb{R})##. This leads to the conclusion that ##\pi
aalma
What is the way to compute ##\pi_1(PGL_2(R))##?
Is it related to defining an action of ##PGL_2(R)## on ##S^3##?

it would be helpful if you can provide me with relevant information regarding this

There is a bundle ##\mathbb{R}^\times\to GL_n(\mathbb{R})\to PGL_n(\mathbb{R})## where ##\mathbb{R}^\times## is the subgroup of nonzero scalar matrices. The identity component of the fiber is contractible, so ##PGL_n(\mathbb{R})## and ##GL_n(\mathbb{R})## has the same homotopy groups in positive degrees, and also the identity component of ##GL_n(\mathbb{R})## is homotopy equivalent to ##SO(n;\mathbb{R})## by performing Gram-Schmidt on the columns.

So in this case, ##\pi_1(PGL_2(\mathbb{R}))\cong\pi_1(SO(2;\mathbb{R}))\cong \pi_1(S^1)\cong\mathbb{Z}.##

Last edited:
malawi_glenn, Euge and aalma
Thanks:)
Is the idea here to move from the fibration you first mentioned to a long exact sequence, knowing that ##\pi_0(GL_2(R))=\pi_0(SO_2(R))##?
When saying "The identity component of the fiber is contractible" to what fiber are you referring and then you mean that ##\pi_1(GL_2(R))=\pi_1(PGL_2(R))##?

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