# Is the singular locus of a variety (as a variety itself) a smooth variety?

A general fact about the singular locus $Sing(X)$ of a variety $X$ (analytic or projective) is that they form a subvariety of the oringinal variety $X$. And we know that the boundary of a manifold have no boundary itself. My simple question is that

## Is $Sing(X)$ (as a variety itself) a smooth variety ?

Intuitively, I can't imagine a picture such that the answer is no.

If the answer is no, another question is that does the singular locus $Sing(X)$ necessarily have dimension less than that of $X$ ?

• If $X$ is the region in $\mathbb{R}^3$ in which all coordinates are non-negative, then the singular set of $X$ is the union of the three non-negative axes, which is not smooth. Jul 29, 2016 at 2:37
• @EduardoLonga That is not an algebraic variety, and I'm not sure what object you're trying to describe (and hence what sense in which that's the singular set).
– user98602
Jul 29, 2016 at 2:41
• I was afraid not.. Jul 29, 2016 at 2:42
• Take the union of three coordinate planes in ${\mathbb C}^3$. If you want an irreducible example, take the quotient of the previous example by the action of the permutation group $S_3$. Jul 29, 2016 at 3:24
• @studiosus , thanks for your example! Jul 29, 2016 at 4:49

• Exactly. ${}{}{}$ Jul 29, 2016 at 5:19
• If $X$ is a given variety, then is there a variety $Y$ whose singular locus is $X$? Jul 29, 2016 at 6:23