1. Affine noncommutative geometry
  2. Free coordinates
  3. 13d. Algebraic geometry

13d. Algebraic geometry

[math] \newcommand{\mathds}{\mathbb}[/math]

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In relation with the general algebraic geometry questions formulated in the beginning of this chapter, and more precisely with the free algebra needed for developing free geometry, we have now at least one clear result on the subject, namely Theorem 13.22. But, long way to go. Modern algebraic geometry is based on modern commutative algebra, as developed by Hilbert, Noether, Zariski and many others, not to talk about Grothendieck and schemes, and such algebraic knowledge is completely lacking in the free setting, preventing for the moment any serious development of free algebraic geometry. This will most likely take a very long time, needing, to start with, a fresh new generation of mathematicians, finding things like Theorem 13.22 trivial, or even lame.


So, forget about modern times, and let us go back instead to the ancient Greeks, with the idea in mind of having some fun with conics. Let us start with:

Definition

A real compact hypersurface in [math]N[/math] variables, denoted [math]X_f\subset\mathbb R^N_+[/math], is the abstract spectrum of a universal [math]C^*[/math]-algebra of the following type,

[[math]] C(X_f)=C^*\left(x_1,\ldots,x_N\Big|x_i=x_i^*,f(x_1,\ldots,x_N)=0\right) [[/math]]
with the noncommutative polynomial [math]f\in\mathbb R \lt x_1,\ldots,x_N \gt [/math] being such the maximal [math]C^*[/math]-norm on the complex [math]*[/math]-algebra [math]\mathbb C \lt x_1,\ldots,x_N \gt /(f)[/math] is bounded.

The boundedness condition above is something quite non-trivial, usually related to tricky operator theory, like sums of squares (SOS) theorems, and so on. If this condition is satisfied, we agree to say that “[math]X_f[/math] exists”. As a first result now, we have:

Theorem

In order for [math]X_f[/math] to exist, the real algebraic manifold

[[math]] X_f^\times=\left\{x\in\mathbb R^N\Big|f(x_1,\ldots,x_N)=0\right\} [[/math]]
must be compact. In addition, in this case we have [math]||x_i||_\times\leq||x_i||[/math], for any [math]i[/math].


Show Proof

Assuming that [math]X_f[/math] exists, our claim is that the algebra of continuous functions on the manifold [math]X_f^\times[/math] in the statement appears from [math]C(X_f)[/math] as follows:

[[math]] C(X_f^\times)=C(X_f)\Big/\Big \lt [x_i,x_j]=0\Big \gt [[/math]]


But this is clear, by applying the Gelfand theorem, and by using as well the Stone-Weierstrass theorem, in order to have arrows in both directions, mapping [math]x_i\to x_i[/math]. With this in hand, we have an embedding of compact quantum spaces, as follows:

[[math]] X_f^\times\subset X_f [[/math]]


The norm estimate is now clear, because such embeddings increase the norms.

■

In practice now, let us first discuss the quadratic case. The existence result here, which is very similar to the one from the classical case, is as follows:

Proposition

Given a quadratic polynomial [math]f\in\mathbb R \lt x_1,\ldots,x_N \gt [/math], written as

[[math]] f=\sum_{ij}A_{ij}x_ix_j+\sum_iB_ix_i+C [[/math]]
the following conditions are equivalent:

  • [math]X_f[/math] exists.
  • [math]X_f^\times[/math] is compact.
  • The symmetric matrix [math]Q=\frac{A+A^t}{2}[/math] is positive or negative.


Show Proof

The implication [math](1)\implies(2)[/math] being known from Theorem 13.24, and the equivalence [math](2)\iff(3)[/math] being well-known, we are left with proving [math](3)\implies(1)[/math]. As a first remark, by applying the adjoint, our manifold [math]X_f[/math] is defined by:

[[math]] \begin{cases} \sum_{ij}A_{ij}x_ix_j+\sum_iB_ix_i+C=0\\ \sum_{ij}A_{ij}x_jx_i+\sum_iB_ix_i+C=0 \end{cases} [[/math]]


In terms of [math]P=\frac{A-A^t}{2}[/math] and [math]Q=\frac{A+A^t}{2}[/math], these equations can be written as:

[[math]] \begin{cases} \sum_{ij}P_{ij}x_ix_j=0\\ \sum_{ij}Q_{ij}x_ix_j+\sum_iB_ix_i+C=0 \end{cases} [[/math]]


Let us first examine the second equation. When regarding [math]x[/math] as a column vector, and [math]B[/math] as a row vector, this equation becomes an equality of [math]1\times1[/math] matrices, as follows:

[[math]] x^tQx+Bx+C=0 [[/math]]


Now let us assume that [math]Q[/math] is positive or negative. Up to a sign change, we can assume [math]Q \gt 0[/math]. We can write [math]Q=UDU^t[/math], with [math]D=diag(d_i)[/math] and [math]d_i \gt 0[/math], and with [math]U\in O_N[/math]. In terms of the vector [math]y=U^tx[/math], and with [math]E=BU[/math], our equation becomes:

[[math]] y^tDy+Ey+C=0 [[/math]]


By reverting back to sums and indices, this equation reads:

[[math]] \sum_id_iy_i^2+\sum_ie_iy_i+C=0 [[/math]]


Now by making squares, this equation takes the following form:

[[math]] \sum_id_i\left(y_i+\frac{e_i}{2d_i}\right)^2=c [[/math]]


By positivity, we deduce that we have the following estimate:

[[math]] \left\|y_i+\frac{e_i}{2d_i}\right\|^2\leq\frac{|c|}{d_i} [[/math]]


Thus our hypersurface [math]X_f[/math] is well-defined, and we are done.

■

We recall that, up to a linear changes of coordinates, there is only one non-trivial compact quadric in [math]\mathbb R^N[/math], namely [math]S^{N-1}_\mathbb R[/math]. In the noncommutative setting the situation is more complicated, because the first equation of [math]X_f[/math] in the above proof, namely [math]\sum_{ij}P_{ij}x_ix_j=0[/math] with [math]P=\frac{A-A^t}{2}[/math], that we have neglected so far, and which is trivial in the classical case, is no longer trivial. By taking into account this equation, we are led to:

Theorem

Up to linear changes of coordinates, the free compact quadrics in [math]\mathbb R^N_+[/math] are the empty set, the point, the standard free sphere [math]S^{N-1}_{\mathbb R,+}[/math], defined by

[[math]] \sum_ix_i^2=1 [[/math]]
and some intermediate spheres [math]S^{N-1}_\mathbb R\subset S\subset S^{N-1}_{\mathbb R,+}[/math], which can be explicitly characterized. Moreover, for all these free quadrics, we have [math]||x_i||=||x_i||_\times[/math], for any [math]i[/math].


Show Proof

We use the computations from the proof of Proposition 13.25. The first equation there, making appear the matrix [math]P=\frac{A-A^t}{2}[/math], is as follows:

[[math]] \sum_{ij}P_{ij}x_ix_j=0 [[/math]]

As for the second equation, up to a linear change of the coordinates, this reads:

[[math]] \sum_iz_i^2=c [[/math]]


At [math]c \lt 0[/math] we obtain the empty set. At [math]c=0[/math] we must have [math]z=0[/math], and depending on whether the first equation is satisfied or not, we obtain either a point, or the empty set. At [math]c \gt 0[/math] now, we can assume by rescaling [math]c=1[/math], and our second equation reads:

[[math]] X_f\subset S^{N-1}_{\mathbb R,+} [[/math]]


As a conclusion, the solutions here are certain subspaces [math]S\subset S^{N-1}_{\mathbb R,+}[/math] which appear via equations of type [math]\sum_{ij}P_{ij}x_ix_j=0[/math], with [math]P\in M_N(\mathbb R)[/math] being antisymmetric, and with [math]x_1,\ldots,x_N[/math] appearing via [math]z_1,\ldots,z_N[/math] via a linear change of variables. Since when redoing the above computation with [math]X_f^\times[/math] at the place of [math]X_f[/math], we obtain [math]X_f=S^{N-1}_\mathbb R[/math], we conclude that our subspaces [math]S\subset S^{N-1}_{\mathbb R,+}[/math] must satisfy:

[[math]] S^{N-1}_\mathbb R\subset S\subset S^{N-1}_{\mathbb R,+} [[/math]]


Thus, we are left with investigating which such subspaces can indeed be solutions. Observe that both the extreme cases can appear as solutions, as shown by:

[[math]] \begin{eqnarray*} X_{2x^2+y^2+\frac{3}{2}xy+\frac{1}{2}yx}&=&S^1_\mathbb R\\ X_{2x^2+y^2+xy+yx}&=&S^1_{\mathbb R,+} \end{eqnarray*} [[/math]]


Finally, the last assertion is clear for the empty set and for the point, and for the remaining hypersurfaces, this follows from [math]S^{N-1}_\mathbb R\subset S\subset S^{N-1}_{\mathbb R,+}[/math].

■

Here is now yet another version of Proposition 13.25, this time by using an opposite idea, namely using as many linear transformations as possible:

Proposition

Given [math]M[/math] real linear functions [math]L_1,\ldots,L_M[/math] in [math]N[/math] noncommuting variables [math]x_1,\ldots,x_N[/math], the following are equivalent:

  • [math]\sum_kL_k(x_1,\ldots,x_N)^2=1[/math] defines a compact hypersurface in [math]\mathbb R^N[/math].
  • [math]\sum_kL_k(x_1,\ldots,x_N)^2=1[/math] defines a compact quantum hypersurface.
  • The matrix formed by the coefficients of [math]L_1,\ldots,L_M[/math] has rank [math]N[/math].


Show Proof

The equivalence [math](1)\iff(2)[/math] follows from [math](1)\iff(2)[/math] in Proposition 13.25, because the surfaces under investigation are quadrics. As for the equivalence [math](2)\iff(3)[/math], this is well-known. More precisely, our equation can be written as:

[[math]] \begin{eqnarray*} 1 &=&\sum_kL_k(x_1,\ldots,x_N)^2\\ &=&\sum_k\sum_iL_{ki}x_i\sum_jL_{kj}x_j\\ &=&\sum_{ij}(L^tL)_{ij}x_ix_j \end{eqnarray*} [[/math]]


Thus, in the context of Proposition 13.25, the underlying square matrix [math]A\in M_N(\mathbb R)[/math] is given by [math]A=L^tL[/math]. It follows that we have [math]Q=A=L^tL[/math], and so the condition [math]Q \gt 0[/math] is equivalent to [math]L^tL[/math] being invertible, and so to [math]L[/math] to have rank [math]N[/math], as claimed.

■

Summarizing, in what concerns the quadrics, the noncommutative theory basically parallels the usual classical theory, with just a few minor twists. In higher degree, however, things look amazingly complicated, because even construcing hypersurfaces via quite trivial sums of squares leads to non-trivial operator theory questions.

General references

Banica, Teo (2024). "Affine noncommutative geometry". arXiv:2012.10973 [math.QA].