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From Wikipedia's morphisms between projective spaces:

Injective linear maps $T \in L(V,W)$ between two vector spaces $V$ and $W$ over the same field $k$ induce mappings of the corresponding projective spaces $P(V) \to P(W)$ via: $$[v] \to [T(v)],$$ where $v$ is a non-zero element of $V$ and [...] denotes the equivalence classes of a vector under the defining identification of the respective projective spaces. Since members of the equivalence class differ by a scalar factor, and linear maps preserve scalar factors, this induced map is well-defined. (If $T$ is not injective, it will have a null space larger than $\{0\}$; in this case the meaning of the class of $T(v)$ is problematic if $v$ is non-zero and in the null space. ...).

  1. In "if $T$ is not injective, the meaning of the class of $T(v)$ is problematic if $v$ is non-zero and in the null space", I wonder what kind of problem that is?
  2. Are morphisms between projective spaces, projective linear transformation, and projective transformation (homography) different names for the same concept?

Thanks and happy holliday!

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    Suppose that $T$ is not injective. Then there exists a nonzero $v$ in $V$ with $T(v) =0$ in $W$. So it has no "class" in $P(W)$. The elements in $P(W)$ are equivalence classes of elements in $W-\{0\}$. Did I misunderstand the question and am I telling you something you already knew?2011-12-24
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    The problem is simply that in projective space there is no origin. If $T(v) = 0 \in W$, then what is the corresponding image of $[v]$ in $P(W)$? If $T$ is injective then $0$ is the only point mapping to $0$, you remove this point on both sides of the map, and everything is fine. If you are thinking of projective space as a manifold or something else then there are certainly non-injective maps: $\mathbf RP^1$, for example, is a circle, and that has a lot of non-injective self-maps.2011-12-24
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    @Ohdur: No, you didn't. Thanks! I forgot that the equivalent classes are in W−{0} not W. Are morphisms between projective spaces, projective linear transformation, and projective transformation (homography) different names for the same concept?2011-12-24
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    @DylanMoreland: Thanks!2011-12-24
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    @Tim. Map $(x,y)$ to $(x+y,x+y)$. This corresponds to the $(2\times 2)$-matrix with 1's everywhere on the standardbasis. This map has a one-dimensional kernel. It is spanned by the vector $(1,-1)$.2011-12-24
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    A projective linear transformation is (in my knowledge) an automorphism of $\mathbf{P}(k^n)$. The group of automorphisms of $\mathbf{P}(k^n)$ is commonly denoted by PGL$_n(k)$. See section 5.3 http://www.math.leidenuniv.nl/~edix/teaching/2010-2011/AG-mastermath/ag.pdf . You can take $k$ to be the field of real numbers or complex numbers if that helps.2011-12-24
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    When you say "a morphism of projective space" do you require it to be an automorphism? If yes, "a morphism of projective space" is a "projective linear transformation". If not, then these concepts are not the same. There are certainly maps from proj space to itself which are not isom's. Dylan Moreland gives an example above.2011-12-24
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    @Ohdur: Thanks! Are projective transformations (homographies) http://en.wikipedia.org/wiki/Homography same as projective linear transformations on a projective space, or morphisms between two projective spaces? By "morphisms between projective spaces", I meant morphisms between two projective spaces which might not be derived from the same vector spaces.2011-12-24

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