# Curvature of a Connection of vector bundle

Let $X$ be a scheme or manifold and $\nabla: V \rightarrow \Omega^1 \otimes V$

be a connection on a vector bundle $V$ on $X$.

Let $R:=\nabla^2$ denote the curvature homomorphism.

Does it hold that $\nabla R =0$?

How does one show this?

-

2
If your $V$ is $TX$ with $X$ a riemannian manifold, and the connection is the Levi-Cività connection, then there is a well known Bianchi identity which would be silly if $\nabla R$ were zero :) Indeed, in this situation, the condition $\nabla R=0$ is the definition of locally symmetric spaces, and not all riemannian manifolds are locally symmetric. – Mariano Suárez-Alvarez♦ Oct 26 '11 at 12:21
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 11, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9159585237503052, "perplexity_flag": "head"}