## If two “homogeneous” algebraic varieties are isomorphic, are they necessarily related by a linear map?

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Let $I,J$ be homogeneous ideals in the algebra of polynomials in $n$ variables over the complex numbers. Let $V(I)$ and $V(J)$ be the affine algebraic varieties that are determined by $I$ and $J$ (not the projective varieties). Suppose that $V(I)$ and $V(J)$ are isomorphic as algebraic varieties. By this I mean that there are polynomial maps $f$ and $g$ from $\mathbb{C}^n$ to itself, such that $f$ restricted to $V(I)$ is a bijection onto $V(J)$, and such that $g$ restricted to $V(J)$ is its inverse.

The question is this: does it follow that there exists a linear map on $\mathbb{C}^n$ that maps $V(I)$ onto $V(J)$?

Thanks to discussions with colleagues (thank you David and Mike), I am quite convinced that if we assume that the origin is the only singular point in $V(I)$ then the answer is yes. Is this true in general?

I think this question is equivalent to the following (see my partial answer below): Is it true that whenever there is an isomorphism between $V(I)$ and $V(J)$, there is also isomorphism that fixes $0$?

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1
The body of your question is not a question, nor is it related to the title as far as I can see! – Mariano Suárez-Alvarez Oct 4 2010 at 17:59
I apologize, the question was saved before I completed typing. It is now a complete question. – Orr Shalit Oct 4 2010 at 18:03

## 4 Answers

Trying to generalize Torsten's answer: It seems that if the cones are isomorphic then the isomorphism can indeed be chosen to preserve the origin.

For a projective variety $V$ let's denote the affine cone by $C(V)$. Torsten says that if $V$ is not a (projective) cone then in $C(V)$ the "cone point" $0$ is the unique point of maximum multiplicity.

$V$ is a projective cone if it is the join of a point $\mathbb P^0$ in the ambient projective space with a variety $W$ in a hyperplane (a hyperplane not containing that point). In this case $C(V)$ is the product of $C(\mathbb P^0)=\mathbb A^1$ and $C(W)$. In the general case $V$ is the join of a linear $\mathbb P^{d-1}$ with some $W$ which is not itself a projective cone, and then $C(V)=C(\mathbb P^{d-1})\times C(W)=\mathbb A^d\times C(W)$. Surely Torsten's statement generalizes to say that the points of maximal multiplicity in $C(V)$ are now those in $\mathbb A^d\times 0$.

So, given $V_1$ and $V_2$ such that $C(V_1)$ and $C(V_2)$ are isomorphic, the two numbers $d_1$ and $d_2$ must be equal, and if the isomorphism does not carry $0$ to $0$ then it can be adjusted to do so using translations in $\mathbb A^d$.

EDIT in response to Orr's comment: Here's what I mean, in your notation. Let $I$ be a homogeneous ideal in $k[x_1,\dots ,x_n]$ corresponding to some "homogeneous" affine variety $V(I)\subset\mathbb A^n$ and let $P(I)\subset \mathbb P^{n-1}$ be the variety that the same ideal defines. (I believe $V(I)$ is called the affine cone on $P(I)$.)

It might happen that after some linear change of coordinates $I$ becomes an ideal generated by a homogeneous ideal $I_1$ in $k[x_1,\dots ,x_p]$ and a homogeneous ideal $I_2$ in $k[x_{p+1},\dots ,x_n]$. If so, then $V(I)$ is the product of $V(I_1)\subset \mathbb A^p$ and $V(I_2)\subset \mathbb A^{n-p}$, and I believe that $P(I)$ would be called the join of the projective varieties $P(I_1)\subset \mathbb P^{p-1}$ and $P(I_2)\subset \mathbb P^{n-p-1}$. In particular if $p=n-1$ and $I_2=0$ then $V(I)=V(I_1)\times \mathbb A^1$ and $P(I)$ is called the projective cone on $P(I_1)$.

Torsten is arguing that if $P(I)$ is not a cone, i.e. if there is no linear change of variable such $I$ is generated by polynomials not involving the last coordinate, then the origin is intrinsically characterized as the unique point in $V(I)$ of maximal multiplicity. I am saying that one can treat the general case in the same way, as follows: Suppose that $P(I)$ is a cone, or a cone on a cone, or ... as far as you can go. That is, make a linear change of variables so that $I$ is generated by polynomials in the first $p$ coordinates with $p$ as small as possible. Thus $V(I)$ is the product of some $V(I_1)$ with $\mathbb A^{n-p}$ and $P(I)$ is the join of the corresponding $P(I_1)$ with $\mathbb P^{n-p-1}$. Now in $V(I)=V(I_1)\times \mathbb A^{n-p}$ the points of $0\times \mathbb A^{n-p}$ are the points of maximum multiplicity, and furthermore any one of them is carried to $0=(0,0)$ by some automorphism of $V(I)$ since $\mathbb A^{n-p}$ has an automorphism group that acts transitively.

The idea of iterated singular locus is not quite so successful. In most cases if the projective variety $S(P(I))$ is $P(J)$ then the homogeneous affine variety $S(V(I))$ will be $V(J)$. In the extreme case when $P(I)$ is smooth, so that $S(P(I))$ is empty, $S(V(I))$ will be $0$, with the exception that if $P(I)$ is a projective space (linearly embedded in $\mathbb P^{n-1}$) then $V(I)$ will be an affine space (linearly embedded and containing the origin) whose singular locus is empty rather than $0$. Thus in the sequence $V(I)$, $S(V(I))$, ... the last nonempty thing will be an affine space, possibly $0$ or possibly bigger. But it will not always be the same thing as before (the maximal affine space such that $V(I)$ is in a linear fashion the product of it with something). For example, if $n=3$ and $P(I)$ is a projective curve which has exactly one singular point but which is not simply the union of lines through that point, then the singular locus of the homogeneous surface $V(I)\subset A^3$ will be a line through the origin but there will be no automorphism of $V(I)$ moving $0$ to another point in that line (or to anything other than $0$).

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Thanks! Let me see if I get this right. You are saying that an affine variety $V = V(I)$ (I homogeneous) is always of the form $\mathbb{A}^d \times C(W)$, where $W$ is a projective variety, and that this $\mathbb{A}^d$ is a geometric invariant. Thus, translations along this space leave $V$ invariant, so we get back the case $0$ goes to $0$. I still do not understand why every $V$ has this form.  Regarding this, another question: one can look at the singular locus of $V$, $S(V)$, and then look at $S(S(V))$ and so on, until you get a subspace. I this subspace the $\mathbb{A}^d$ from above? – Orr Shalit Oct 9 2010 at 12:51

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Yes, assuming at least that the cone point is the only singular point. Hence any isomorphism will preserve the ideal of that point which is the ideal of elements of positive degree.(I think that the cone should in some sense be the most singular point in general and hence would still be preserved.) You can think of the isomorphism and its inverse as a graded isomorphism where the coordinate rings are graded by the powers of the ideal of the cone point. They then give graded isomorphisms of the associated graded rings. These associated graded rings are however the original coordinate rings. Hence we get a graded isomorphism of the coordinat rings and these isomorphisms are equal to those induced by the linear maps on the degree $1$ part.

Addendum: You can considerably weaken the condition that the cone point is the only singular point. Assume that the associated projective scheme to he ideals are varieties which are not cones. Then the multiplicity of any point outside of the cone point is equal to the multiplicity of the image point on the projective variety and that multiplicity is smaller than the degree of the variety. That degree however is just the multiplicity of the cone point. Hence, the cone points are the points of maximum multiplicity on the respective variety and are therefore taken to each other by an isomorphism. It seems likely that the case of the projective varieties being cones can be further analysed.

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Thank you for the answer. I understand the first part. It would be interesting to know if the requirement that the cone point be the only the only singular point be removed completely.   I do not understand the addendum - is there perhaps an elementary explanation?  – Orr Shalit Oct 4 2010 at 22:55

A (projective) line is isomorphic to a conic, but not via a linear map. Do you mean to ask that two projective varieties whose degrees are equal would be isomorphic via a linear map? That still does not seem to be true: take a variety with two non-linearly equivalent very ample divisor of the same degree. The embeddings by the complete linear systems of these two divisors give isomorphic varieties (you can easily make them live in the same projective space) but they cannot be mapped into one another by a linear map, because that would imply the two original divisors being linearly equivalent.

The above condition should be easy to satisfy as soon as the Picard number is larger than $1$.

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This answer was posted before the correction to the question... – Sándor Kovács Oct 4 2010 at 18:17

I'll share here two proofs of the fact I am interested in under the assumption that the origin is the only singular point.

Proof 1: (This is equivalent I think to Torsten's answer, and was a explained to me by a colleagues - thank you David and Mike) Let $F: V(I) \rightarrow V(J)$ be an isomorphism. Then since $0$ is the only singular point in $V(I)$, then it is mapped to $0 \in V(J)$. Now the derivative of $F$ at $0$, call it $DF$, is a linear map in $\mathbb{C}^n$ that takes the tangent cone of $V(I)$ at $0 \in V(I)$ to the tangent cone of $V(J)$ at $0 \in V(J)$. But these tangent cones are $V(I)$ and $V(J)$, respectively. Thus, $DF$ is the required linear map taking $V(I)$ onto $V(J)$.

Proof 2: There are some technical details missing here. Let $F: V(I) \rightarrow V(J)$ be an isomorphism. Again, $F$ must take $0$ to $0$, under the assumption that $0$ is the only singular point. Thus, $F$ has the form $$F(z) = Az + \textrm{ higher order terms .}$$ Now define $F_t$ by $$F_t(z) = tF(z/t) .$$ Since $I$ and $J$ are homogeneous, $V(I)$ and $V(J)$ are invariant under scalings, so $F_t$ is again an isomorphismof $V(I)$ and $V(J)$. But $F_t$ has the form $$F_t(z) = Az + \frac{1}{t}(\textrm{higher order terms}).$$ Taking $t \rightarrow \infty$, we converge to the isomorphism (hopefully) $z \mapsto Az$.

That's for the case when $0$ is the only singular point. In fact, all that is used is that there is an isomorphism taking $0$ to $0$. Is it true that whenever there is an isomorphism between $V(I)$ and $V(J)$, there is also isomorphism that fixes $0$? (here $I$ and $J$ are assumed homogeneous, of course).

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