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Callum
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I think there must be some different ideas of what your question means, looking at the comments. I understood it to mean as in the decomposition of tensor powers of a $\mathfrak{g}$-representation into irreducible subrepresentations. In which case, the answer is a definite yes. For a start, the action of $\mathfrak{g}$ on any symmetric power of a representation preserves the degree. So in this way $S(\mathfrak{g}) = \bigoplus_{i=0}^\infty S^{i}(\mathfrak{g})$. Then the isotypic components in this sense just refers to the decomposition of these finite dimensional pieces into irreducibles which are automatically finite dimensional themselves. The explicit decompositions themselves can be quite hard to work out although there are formulae for these things. This is the kind of thing that "plethysm" refers to.

The invariant polynomials in this view form trivial subrepresentations but they would not be regarded as a single isotypic component (indeed they appear at various degrees and there are more than $\mathrm{rank}\ \mathfrak{g}$ of them, those are just the generators)

Edit: having looked through the edits to the question I understand the confusion a little better. You originally asked about whether copies of each irreducible appeared finitely many times. I would hesitate to call these isotypic components but the answer here is no. As pointed out in the comments the collection of all trivial subreps form an infinite-dimensional component. This is, in essence, the span of all polynomials generated by the Casimir elements.

I think there must be some different ideas of what your question means, looking at the comments. I understood it to mean as in the decomposition of tensor powers of a $\mathfrak{g}$-representation into irreducible subrepresentations. In which case, the answer is a definite yes. For a start, the action of $\mathfrak{g}$ on any symmetric power of a representation preserves the degree. So in this way $S(\mathfrak{g}) = \bigoplus_{i=0}^\infty S^{i}(\mathfrak{g})$. Then the isotypic components in this sense just refers to the decomposition of these finite dimensional pieces into irreducibles which are automatically finite dimensional themselves. The explicit decompositions themselves can be quite hard to work out although there are formulae for these things. This is the kind of thing that "plethysm" refers to.

The invariant polynomials in this view form trivial subrepresentations but they would not be regarded as a single isotypic component (indeed they appear at various degrees and there are more than $\mathrm{rank}\ \mathfrak{g}$ of them, those are just the generators)

I think there must be some different ideas of what your question means, looking at the comments. I understood it to mean as in the decomposition of tensor powers of a $\mathfrak{g}$-representation into irreducible subrepresentations. In which case, the answer is a definite yes. For a start, the action of $\mathfrak{g}$ on any symmetric power of a representation preserves the degree. So in this way $S(\mathfrak{g}) = \bigoplus_{i=0}^\infty S^{i}(\mathfrak{g})$. Then the isotypic components in this sense just refers to the decomposition of these finite dimensional pieces into irreducibles which are automatically finite dimensional themselves. The explicit decompositions themselves can be quite hard to work out although there are formulae for these things. This is the kind of thing that "plethysm" refers to.

The invariant polynomials in this view form trivial subrepresentations but they would not be regarded as a single isotypic component (indeed they appear at various degrees and there are more than $\mathrm{rank}\ \mathfrak{g}$ of them, those are just the generators)

Edit: having looked through the edits to the question I understand the confusion a little better. You originally asked about whether copies of each irreducible appeared finitely many times. I would hesitate to call these isotypic components but the answer here is no. As pointed out in the comments the collection of all trivial subreps form an infinite-dimensional component. This is, in essence, the span of all polynomials generated by the Casimir elements.

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Callum
  • 954
  • 6
  • 15

I think there must be some different ideas of what your question means, looking at the comments. I understood it to mean as in the decomposition of tensor powers of a $\mathfrak{g}$-representation into irreducible subrepresentations. In which case, the answer is a definite yes. For a start, the action of $\mathfrak{g}$ on any symmetric power of a representation preserves the degree. So in this way $S(\mathfrak{g}) = \bigoplus_{i=0}^\infty S^{i}(\mathfrak{g})$. Then the isotypic components in this sense just refers to the decomposition of these finite dimensional pieces into irreducibles which are automatically finite dimensional themselves. The explicit decompositions themselves can be quite hard to work out although there are formulae for these things. This is the kind of thing that "plethysm" refers to.

The invariant polynomials in this view form trivial subrepresentations but they would not be regarded as a single isotypic component (indeed they appear at various degrees and there are more than $\mathrm{rank}\ \mathfrak{g}$ of them, those are just the generators)

I think there must be some different ideas of what your question means, looking at the comments. I understood it to mean as in the decomposition of tensor powers of $\mathfrak{g}$-representation. In which case, the answer is a definite yes. For a start, the action of $\mathfrak{g}$ on any symmetric power of a representation preserves the degree. So in this way $S(\mathfrak{g}) = \bigoplus_{i=0}^\infty S^{i}(\mathfrak{g})$. Then the isotypic components in this sense just refers to the decomposition of these finite dimensional pieces into irreducibles which are automatically finite dimensional themselves. The explicit decompositions themselves can be quite hard to work out although there are formulae for these things.

I think there must be some different ideas of what your question means, looking at the comments. I understood it to mean as in the decomposition of tensor powers of a $\mathfrak{g}$-representation into irreducible subrepresentations. In which case, the answer is a definite yes. For a start, the action of $\mathfrak{g}$ on any symmetric power of a representation preserves the degree. So in this way $S(\mathfrak{g}) = \bigoplus_{i=0}^\infty S^{i}(\mathfrak{g})$. Then the isotypic components in this sense just refers to the decomposition of these finite dimensional pieces into irreducibles which are automatically finite dimensional themselves. The explicit decompositions themselves can be quite hard to work out although there are formulae for these things. This is the kind of thing that "plethysm" refers to.

The invariant polynomials in this view form trivial subrepresentations but they would not be regarded as a single isotypic component (indeed they appear at various degrees and there are more than $\mathrm{rank}\ \mathfrak{g}$ of them, those are just the generators)

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Callum
  • 954
  • 6
  • 15

I think there must be some different ideas of what your question means, looking at the comments. I understood it to mean as in the decomposition of tensor powers of $\mathfrak{g}$-representation. In which case, the answer is a definite yes. For a start, the action of $\mathfrak{g}$ on any symmetric power of a representation preserves the degree. So in this way $S(\mathfrak{g}) = \bigoplus_{i=0}^\infty S^{i}(\mathfrak{g})$. Then the isotypic components in this sense just refers to the decomposition of these finite dimensional pieces into irreducibles which are automatically finite dimensional themselves. The explicit decompositions themselves can be quite hard to work out although there are formulae for these things.