I am studying some algebraic geometry and surfaces of Riemann and reading some books... often appears an expression whose meaning is not clear to me. So I would like some help to understand this. The expression in question is "intrinsically".

I have separated some of the moments in which I have found (and continue) with doubts about what does it mean meant by the use of the word "intrinsically".

1) Nevertheless, we have, in fact, gained much, for we have defined multiplicities of $(d — 1)$-dimensional varieties intrinsically, in terms of properties of a ring of the function field of $V$, and this will be found to be of immense value in later

2) Recall that if $M=\mathbb{C}^g/\Lambda$ is any complex torus, then the tangent spaces $\{T'_{\lambda}(M) \}_{\lambda\in M} $ are all naturally identified with $\mathbb{C}^g$. Thus if $X\subset M$ is any analytic subvariety of dimension $k$, $X^* = X — X_{\text{sing}}$ the smooth locus of $X$, we can define the Gauss map $$ \begin{array}{llll} {\cal{G}}_ {X} :&X^*&\longrightarrow&G(k, g) \\ &p&\longmapsto&T'_{p}(X)\subset T'_{p}(M)= \mathbb{C}^g.\\ \end{array} $$

We see immediately that $\cal G$ is intrinsically defined and that it does not vary if $X$ is translated in $M$.

So that's my question. What do authors mean by intrinsically in these two passages.

Thank You!


1 Answer 1


In (algebraic/differential) geometry an intrinsic property of a variety $X$ usually means one that does not depend on an embedding $X\hookrightarrow M$ into some larger variety.

For example, the properties of an affine variety that depend only on the ring of regular functions rather than some specific ideal.

The tangent bundle of a manifold $X$ is an intrinsic object and the normal bundle of $X\subset M$ is an extrinsic object.

I have to mention Gauss' Theorema Egregium which is called that way (by Gauss himself) because of the egregium (remarkable) fact that the Gaussian curvature is an intrinsic invariant despite it is defined extrinsically.


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