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$SO(n)$'s rep $\rho: SO( n) \to SL(d,\mathbf{R})$ for some dimension d of representations (reps).

The projective group $PSO(n)$'s rep $\rho'$ must be a rep of $SO(n)$'s rep $\rho$. But $SO( n)$'s rep $\rho$ must be a rep of $PSO( n)$'s rep $\rho'$.

But what are the constraints of the $\rho'$ such that which $\rho$ are not survived in $\rho'$?

  • Does the adjoint rep $(n(n-1)/2)$-dim rep of $SO(n )$ survive in $PSO( n)$?

  • Does the vector rep ($n$-dim rep) of $SO(n )$ not survive in $PSO( n)$?

  • What are the rules of the general reps of $SO( n)$ that survived or not in $PSO( n)$?

Марина Marina S
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    Just make sure the $SO(n)$-representation has kernel containing $Z(SO(n))$ and you will get a representation of $PSO(n)$. – user10354138 Sep 12 '21 at 13:54
  • Really? Even for defining representation of $SO(n)$ for all $n$? – user10354138 Sep 12 '21 at 15:15
  • I meant to say, do you mean that "the ()-representation :()→(,) , has kernel of this map, containing (()) in ())?" – Марина Marina S Sep 12 '21 at 15:32
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    Yes. The representation factors through PSO(n) (i.e., $\rho$ is $SO(n)\xrightarrow{\pi}PSO(n)\xrightarrow{\bar\rho} SL(d,\mathbb{R})$) iff $\ker\rho\geq\ker\pi=Z(SO(n))$, which is a special case of the universal property of quotients. – user10354138 Sep 12 '21 at 15:39
  • Related: https://math.stackexchange.com/q/3108122/96384, https://math.stackexchange.com/q/347459/96384, https://math.stackexchange.com/q/3762807/96384. – Torsten Schoeneberg Sep 13 '21 at 04:22
  • One thing to note is that the adjoint rep should always work. Any Lie group with a given Lie algebra has an adjoint representation. Under some basic assumptions (algebraically closed I think is all we need) these are all the same rep. – Callum Sep 18 '21 at 22:28

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