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respect --> respects
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Matthieu Romagny
Matthieu Romagny
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I don't think so. Consider the case which should be the most difficult to split canonically, the case when the $p$'th power map is zero. The automorphism group is then equal to the linear automorphism group of $\mathfrak g$ and I assume further that $k$ is an infinite field and $\mathfrak g$ a finite dimensional vector space. I then interpret "canonical" as saying in particular that a canonical splitting respectrespects the action of the automorphism group, the general linear group of $\mathfrak g$. As the relations in the restricted enveloping algebra are of the form $x^p=0$ everything is graded and the grading can be read off from how scalar multiplication acts. As $k$ is infinite this means that a canonical splitting must be homogeneous. This implies that the map from $\mathrm{Sym}^p\mathfrak g$ onto the degree $p$ part of $U^{[p]}(\mathfrak g)$ must split equivariantly. However, the kernel of this map is the image of $\mathfrak g^{(p)}$ given by the $p$'th power map in the symmetric algebra and it is well-known that that inclusion does not split as a map of representations of the general linear group.

I don't think so. Consider the case which should be the most difficult to split canonically, the case when the $p$'th power map is zero. The automorphism group is then equal to the linear automorphism group of $\mathfrak g$ and I assume further that $k$ is an infinite field and $\mathfrak g$ a finite dimensional vector space. I then interpret "canonical" as saying in particular that a canonical splitting respect the action of the automorphism group, the general linear group of $\mathfrak g$. As the relations in the restricted enveloping algebra are of the form $x^p=0$ everything is graded and the grading can be read off from how scalar multiplication acts. As $k$ is infinite this means that a canonical splitting must be homogeneous. This implies that the map from $\mathrm{Sym}^p\mathfrak g$ onto the degree $p$ part of $U^{[p]}(\mathfrak g)$ must split equivariantly. However, the kernel of this map is the image of $\mathfrak g^{(p)}$ given by the $p$'th power map in the symmetric algebra and it is well-known that that inclusion does not split as a map of representations of the general linear group.

I don't think so. Consider the case which should be the most difficult to split canonically, the case when the $p$'th power map is zero. The automorphism group is then equal to the linear automorphism group of $\mathfrak g$ and I assume further that $k$ is an infinite field and $\mathfrak g$ a finite dimensional vector space. I then interpret "canonical" as saying in particular that a canonical splitting respects the action of the automorphism group, the general linear group of $\mathfrak g$. As the relations in the restricted enveloping algebra are of the form $x^p=0$ everything is graded and the grading can be read off from how scalar multiplication acts. As $k$ is infinite this means that a canonical splitting must be homogeneous. This implies that the map from $\mathrm{Sym}^p\mathfrak g$ onto the degree $p$ part of $U^{[p]}(\mathfrak g)$ must split equivariantly. However, the kernel of this map is the image of $\mathfrak g^{(p)}$ given by the $p$'th power map in the symmetric algebra and it is well-known that that inclusion does not split as a map of representations of the general linear group.

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Torsten Ekedahl
Torsten Ekedahl
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I don't think so. Consider the case which should be the most difficult to split canonically, the case when the $p$'th power map is zero. The automorphism group is then equal to the linear automorphism group of $\mathfrak g$ and I assume further that $k$ is an infinite field and $\mathfrak g$ a finite dimensional vector space. I then interpret "canonical" as saying in particular that a canonical splitting respect the action of the automorphism group, the general linear group of $\mathfrak g$. As the relations in the restricted enveloping algebra are of the form $x^p=0$ everything is graded and the grading can be read off from how scalar multiplication acts. As $k$ is infinite this means that a canonical splitting must be homogeneous. This implies that the map from $\mathrm{Sym}^p\mathfrak g$ onto the degree $p$ part of $U^{[p]}(\mathfrak g)$ must split equivariantly. However, the kernel of this map is the image of $\mathfrak g^{(p)}$ given by the $p$'th power map in the symmetric algebra and it is well-known that that inclusion does not split as a map of representations of the general linear group.