Irreducible representation of GL(D)

[gentsagree]

Hi,

I'm reading: "Let $v_{a}$ represent a generic element of $R^{D}$. The action of a non-singular linear operator on this space gives a D-dimensional irreducible representation V of GL(D); indeed, this representation defines the group itself".

I have a couple of questions:

1. How do I know that the rep will be IRREDUCIBLE? Is it a straightforward consequence of the linearity of the operators, or otherwise?

2. What does the last bit mean? Is it that the representation furnished by the action of linear ops on R is the "fundamental" of GL(D)?

Thanks

 

[jvicens]

for your help!

Hi there,

As a scientist, I can provide some insight into your questions.

1. The fact that the representation V is irreducible is a direct consequence of the linearity of the operators. This is because the linearity implies that the representation cannot be decomposed into smaller, independent representations. In other words, the action of the linear operators on R^D is a complete and irreducible representation of GL(D).

2. The last bit means that the representation V defines the group GL(D) itself. In other words, the representation V captures all the essential properties of the group GL(D) and can be used to fully describe and understand the group. This is because the action of the linear operators on R^D is a faithful representation of GL(D), meaning that it preserves all the group operations and structure.

I hope this helps clarify the concept for you. Let me know if you have any further questions.