ncg

commit 8a865911b1bb0ce40ff8628e4d4401d888f7edd1
parent 5dc97ef338dff58db6d6d5bcce873c05e329e8f5
Author: miksa <milutin@popovic.xyz>
Date:   Tue, 11 May 2021 13:27:47 +0200

fixed week5

Diffstat:
M
src/week5.tex
 | 
17
+++++++++++------

1 file changed, 11 insertions(+), 6 deletions(-)
diff --git a/src/week5.tex b/src/week5.tex
@@ -281,7 +281,7 @@ exercises.
 
 \begin{definition}
     Given an finite spectral triple $(A, H, D)$, the $A$-bimodule of
-    Connes' differential one form is:
+    Connes' differential one-forms is:
     \begin{equation}
         \Omega _D ^1 (A) := \left\{ \sum _k a_k[D, b_k]: a_k, b_k \in A \right\}
     \end{equation}
@@ -453,8 +453,14 @@ a finite spectral triple on $B$, $(B, H', D')$
 \begin{equation}
     H' = E \otimes _A H
 \end{equation}
-This extends the left action on $B$ with the right action and inherits the
-$\mathbb{C}$ valued inner product space.
+We might define $D'$ with $D'(e \otimes \xi) = e\otimes D\xi$, thought this
+would not satisfy the ideal defining the balanced tensor product over $A$,
+which is generated by elements of the form
+\begin{align}
+    e a \otimes \xi - e\otimes a \xi ;\;\;\;\; e\in E, a\in A, \xi \in H
+\end{align}
+This inherits the left action on $B$ from $E$ and has a $\mathbb{C}$
+valued inner product space. $B$ also satisfies the ideal.
 \begin{equation}
     D'(e\otimes \xi) = e \otimes D \xi + \nabla (e) \xi \;\;\;\; e\in
     E, a\in A
@@ -465,14 +471,13 @@ associated with the  derivation $d=[D, \cdot]$ and satisfying the
 \begin{equation}
     \nabla(ae) = \nabla(e)a + e \otimes [D, a] \;\;\;\;\;  e\in E,\; a\in A
 \end{equation}
-Then the linearity of the balanced tensor product $E \otimes _A H$ is
-satisfied
+Then $D'$ is well defined on $E \otimes _A H$:
 \begin{align*}
     D'(ea \otimes \xi - e \otimes a \xi) &=  D'(ea \otimes \xi) - D'(e
     \otimes \xi) \\
     &= ea\otimes D\xi + \nabla(ae) \xi - e \otimes D(a\xi ) - \nabla (e)a
     \xi \\
-    &= 0
+    &= 0.
 \end{align*}
 With the information thus far we can prove the following theorem
 \begin{theorem}